Differential Equations Class 12th Mathematics Part Ii CBSE Solution

Class 12^th Mathematics Part Ii CBSE Solution

Exercise 9.1

1. d^4y/dx^4 + sin (y^there there eξ sts there there eξ sts there there eξ sts) = 0…
2. y′ + 5y = 0 Determine order and degree (if defined) of differential equations…
3. (ds/dt)^4 + 3s d^2s/dt^2 = 0 Determine order and degree (if defined) of…
4. (d^2y/dx^2)^2 + cos (dy/dx) = 0 Determine order and degree (if defined) of…
5. d^2y/dx^2 = cos3x+sin3x Determine order and degree (if defined) of differential…
6. (y″′)^2 + (y″)^3 + (y′)^4 + y^5 = 0 Determine order and degree (if defined) of…
7. y″′ +2y” + y’ = 0 Determine order and degree (if defined) of differential…
8. y′ + y = ex Determine order and degree (if defined) of differential equations…
9. y″ + (y′) + 2y = 0 Determine order and degree (if defined) of differential…
10. y″ + 2y′ + sin y = 0 Determine order and degree (if defined) of differential…
11. The degree of the differential equation (d^2y/dx^2)^3 + (dy/dx)^2 + sin (dy/dx)…
12. The order of the differential equation 2x^2 d^2y/dx^2 - 3 dy/dx + y = 0 isA. 2…

Exercise 9.2

1. y = ex + 1 : y′′ - y′ = 0 In each of the question verify that the given…
2. y = x^2 + 2x + C : y′ - 2x - 2 = 0 In each of the question verify that the given…
3. y = cos x + C : y′ + sin x = 0 In each of the question verify that the given…
4. y = root 1+x^2 : y^there there eξ sts = xy/1+x^2 In each of the question verify…
5. y = Ax : xy′ = y (x ≠ 0) In each of the question verify that the given functions…
6. y = x sin x : xy′ = y + x root x^2 - y^2 (x ≠ 0 and x y or x - y) In each of the…
7. xy = log y + C : y^there there eξ sts = y^2/1-xy (xy not equal 1) In each of the…
8. y - cos y = x : (y sin y + cos y + x)y′ = y In each of the question verify that…
9. x + y = tan-1y : y^2 y′ + y^2 + 1 = 0 In each of the question verify that the…
10. y = root a^2 - x^2 x in (- a , a) :x+y dy/dx = 0 (y not equal 0) In each of the…
11. The number of arbitrary constants in the general solution of a differential…
12. The number of arbitrary constants in the particular solution of a differential…

Exercise 9.3

1. x/a + y/b = 1 In each of the question, form a differential equation representing…
2. y^2 = a(b^2 - x^2) In each of the question, form a differential equation…
3. y = a e3x + b e-2x In each of the question, form a differential equation…
4. y = e2x (a + bx) In each of the question, form a differential equation…
5. y = ex (a cos x + b sin x) In each of the question, form a differential equation…
6. Form the differential equation of the family of circles touching the y-axis at…
7. Form the differential equation of the family of parabolas having vertex at…
8. Form the differential equation of the family of ellipses having foci on y-axis…
9. Form the differential equation of the family of hyperbolas having foci on x-axis…
10. Form the differential equation of the family of circles having centre on y-axis…
11. Which of the following differential equations has y = c1ex + c2e-x as the…
12. Which of the following differential equations has y = x as one of its…

Exercise 9.4

1. dy/dx = 1-cosx/1+cosx For each of the differential equations in question, find…
2. dy/dx = root 4-y^2 (-2y2) For each of the differential equations in question,…
3. dy/dx + y = 1 (y not equal 1) For each of the differential equations in…
4. sec^2 x tan y dx + sec^2 y tan x dy For each of the differential equations in…
5. (ex + e-x)dy - (ex - e-x)dx = 0 For each of the differential equations in…
6. dy/dx = (1+x^2) (1+y^2) For each of the differential equations in question, find…
7. y log y dx - x dy = 0 For each of the differential equations in question, find…
8. x^5 dy/dx = - y^5 For each of the differential equations in question, find the…
9. dy/dx = sin^-1x For each of the differential equations in question, find the…
10. extan y dx + 1(1 - ex)sec^2 y dy = 0 For each of the differential equations in…
11. (x^3 + x^2 + x + 1) dy/dx = 2x^2 + x, y = 1 when x = 0 For each of the…
12. x(x^2 - 1) dy/dx = 1; y = 0 when x = 0 For each of the differential equations…
13. cos dy/dx = a; y = 2 when x = 0 For each of the differential equations in…
14. dy/dx = y tan x; y = 1 when x = 0 For each of the differential equations in…
15. Find the equation of a curve passing through the point (0, 0) and whose…
16. For the differential equation xy dy/dx = (x+2) (y+2) find the solution curve…
17. Find the equation of a curve passing through the point (0, -2) given that at…
18. At any point (x, y) of a curve, the slope of the tangent is twice the slope of…
19. The volume of spherical balloon being inflated changes at a constant rate. If…
20. In a bank, principal increases continuously at the rate of r% per year. Find…
21. In a bank, principal increases continuously at the rate of 5% per year. An…
22. In a culture, the bacteria count is 1,00,000. The number is increased by 10% in…
23. The general solution of the differential equation dy/dx = e^x+y isA. ex + e-y =…

Exercise 9.5

1. (x^2 + xy)dy = (x^2 + y^2)dx In each of the question, show that the given…
2. y^there there eξ sts = x+y/x In each of the question, show that the given…
3. (x - y)dy - (x + y)dx = 0 In each of the question, show that the given…
4. (x^2 - y^2)dx + 2xy dy = 0 In each of the question, show that the given…
5. x^2 dy/dx = x^2 - 2y^2 + xy In each of the question, show that the given…
6. xdy-ydx = root x^2 + y^2 dx In each of the question, show that the given…
7. xcos (y/x) + ysin (y/x) ydx = ysin (y/x) - xcos (y/x) xdy In each of the…
8. x dy/dx - y+xsin (y/x) = 0 In each of the question, show that the given…
9. ydx+xlog (y/x) dy-2xdy = 0 In each of the question, show that the given…
10. (1+e^x/y) dx+e^x/y (1 - x/y) dy = 0 In each of the question, show that the…
11. (x + y)dy + (x - y) dx = 0; y = 1 when x = 1 For each of the differential…
12. x^2 dy + (xy + y^2)dx = 0; y = 1 when x = 1 For each of the differential…
13. [xsin^2 (y/x) - y]dx+xdy = 0 y = pi /4 whenx = 1 For each of the differential…
14. dy/dx - y/x + cosec (y/x) = 0 y = 0 when x = 1 For each of the differential…
15. 2xy+y^2 - 2x^2 dy/dx = 0 y = 2whenx = 1 For each of the differential equations…
16. A homogeneous differential equation of the from dx/dy = h (x/y) can be solved…
17. A homogeneous differential equation of the from dx/dy = h (x/y) can be solved…

Exercise 9.6

1. dy/dx + 2y = sinx For each of the differential equations given in question, find…
2. dy/dx + 3y = e^-2x For each of the differential equations given in question,…
3. dy/dx + y/x = x^2 For each of the differential equations given in question, find…
4. dy/dx + (secx) y = tanx (0 less than equal to x pi /2) For each of the…
5. cos^2x dy/dx + y = tanx (0 less than equal to x pi /2) For each of the…
6. x dy/dx + 2y = x^2logx For each of the differential equations given in question,…
7. xlogx dy/dx + y = 2/x logx For each of the differential equations given in…
8. (1 + x^2)dy + 2xy dx = cot x dx (x ≠ 0) For each of the differential equations…
9. x dy/dx + y-x+xycotx = 0 (x not equal 0) For each of the differential equations…
10. (x+y) dy/dx = 1 For each of the differential equations given in question, find…
11. y dx + (x - y^2)dy = 0 For each of the differential equations given in…
12. (x+3y^2) dy/dx = y (y0) For each of the differential equations given in…
13. dy/dx + 2ytanx = sinxy = 0 x = pi /3 For each of the differential equations…
14. (1+x^2) dy/dx + 2xy = 1/1+x^2 y = 0 x = 1 For each of the differential…
15. dy/dx - 3ycotx = sin2xy = 2 x = pi /2 For each of the differential equations…
16. Find the equation of a curve passing through the origin given that the slope of…
17. Find the equation of a curve passing through the point (0, 2) given that the…
18. The Integrating Factor of the differential equation x dy/dx - y = 2x^2 isA. e-x…
19. The Integrating Factor of the differential equation (1-y^2) dx/dy + yx = ay…

Miscellaneous Exercise

1. d^2y/dx^2 + 5x (dy/dx)^2 - 6y = logx For each of the differential equations…
2. (dy/dx)^3 - 4 (dy/dx)^2 + 7y = sinx For each of the differential equations…
3. d^4y/dx^4 - sin (d^3y/dx^3) = 0 For each of the differential equations given…
4. xy = a ex + b e-x + x^2 : x d^2y/dx^2 + 2 dy/dx - xy+x^2 - 2 = 0 For each of…
5. y = ex (a cos x + b sin x) : d^2y/dx^2 - 2 dy/dx + 2y = 0 For each of the…
6. y = x sin 3x : d^2y/dx^2 + 9y-6cos3x = 0 For each of the exercises given below,…
7. x^2 = 2y^2 log y : (x^2 + y^2) dy/dx - xy = 0 For each of the exercises given…
8. Form the differential equation representing the family of curves given by (x -…
9. Prove that x^2 - y^2 = c (x^2 + y^2)^2 is the general solution of differential…
10. Form the differential equation of the family of circles in the first quadrant…
11. Find the general solution of the differential equation dy/dx + root 1-y^2/1-x^2…
12. Show that the general solution of the differential equation dy/dx + y^2 +…
13. Find the equation of the curve passing through the point (0 , pi /4) whose…
14. Find the particular solution of the differential equation (1 + e2x) dy + (1 +…
15. Solve the differential equation ye^x/y dx = (xe^x/y + y^2) dy (y not equal 0)…
16. Find a particular solution of the differential equation (x - y) (dx + dy) = dx…
17. Solve the differential equation [e^- 2 root x/root x - y/root x] dx/dy = 1 (x…
18. Find a particular solution of the differential equation dy/dx + ycotx =…
19. Find a particular solution of the differential equation (x+1) dy/dx = 2e^-y - 1…
20. The population of a village increases continuously at the rate proportional to…
21. The general solution of the differential equation ydx-xdx/y = 0 isA. xy = C B.…
22. The general solution of a differential equation of the type dx/dy + p_1x = q_1…
23. The general solution of the differential equation ex dy + (y ex + 2x) dx = 0…

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Exercise 9.1

Question 1.

Determine order and degree (if defined) of differential equations given

Answer:

It is given that equation is 

⇒ y”” + sin(y’’’) = 0

We can see that the highest order derivative present in the differential is y””.

Thus, its order is four. The given differential equation is not a polynomial equation in its derivative.

Therefore, its degree is not defined.

Question 2.

Determine order and degree (if defined) of differential equations given

y′ + 5y = 0

Answer:

It is given that equation is y’ + 5y = 0

We can see that the highest order derivative present in the differential is y’.

Thus, its order is one. It is polynomial equation in y’. The highest power raised to y’ is 1.

Therefore, its degree is one.

Question 3.

Determine order and degree (if defined) of differential equations given

Answer:

It is given that equation is 

We can see that the highest order derivative present in the given differential equation is

Thus, its order is two. It is polynomial equation in  and 

Therefore, its degree is one.

Question 4.

Determine order and degree (if defined) of differential equations given

Answer:

It is given that equation is 

We can see that the highest order derivative present in the given differential equation is.

Thus, its order is two. The given differential equation is not a polynomial equation in its derivative.

Therefore, its degree is not defined.

Question 5.

Determine order and degree (if defined) of differential equations given

Answer:

It is given that equation is 

We can see that the highest order derivative present in the given differential equation is

Thus, its order is two. It is polynomial equation in  and the power is 1.

Therefore, its degree is one.

Question 6.

Determine order and degree (if defined) of differential equations given

(y″′)² + (y″)³ + (y′)⁴ + y⁵ = 0

Answer:

It is given that equation is (y”’)² + (y”)³ +(y’)⁴ + y⁵ = 0

We can see that the highest order derivative present in the differential is y”’.

Thus, its order is three. It is polynomial equation in y”’, y” and y’

So, the highest power raised to y”’ is 2.

Therefore, its degree is two.

Question 7.

Determine order and degree (if defined) of differential equations given

y″′ +2y” + y’ = 0

Answer:

It is given that equation is y″′ +2y” + y’ = 0

We can see that the highest order derivative present in the differential is y”’.

Thus, its order is three. It is polynomial equation in y”’, y” and y’

So, the highest power raised to y”’ is 1.

Therefore, its degree is one.

Question 8.

Determine order and degree (if defined) of differential equations given

y′ + y = e^x

Answer:

It is given that equation is y’ + y = e^x

⇒ y’ + y – e^x = 0

We can see that the highest order derivative present in the differential is y’.

Thus, its order is one. It is polynomial equation in y’

So, the highest power raised to y’ is 1.

Therefore, its degree is one.

Question 9.

Determine order and degree (if defined) of differential equations given

y″ + (y′) + 2y = 0

Answer:

It is given that equation is y’’ + (y’)² + 2y = 0

We can see that the highest order derivative present in the differential is y”.

Thus, its order is two. It is polynomial equation in y” and y’

So, the highest power raised to y” is 1.

Therefore, its degree is one.

Question 10.

Determine order and degree (if defined) of differential equations given

y″ + 2y′ + sin y = 0

Answer:

It is given that equation is y’’ + 2y’ + siny = 0

We can see that the highest order derivative present in the differential is y”.

Thus, its order is two. It is polynomial equation in y” and y’

So, the highest power raised to y” is 1.

Therefore, its degree is one.

Question 11.

The degree of the differential equation

 is
A. 3

B. 2

C. 1

D. not defined

Answer:

It is given that equation is 

We can see that the highest order derivative present in the given differential equation is.

Thus, its order is three.

The given differential equation is not a polynomial equation in its derivative.

Therefore, its degree is not defined.

Question 12.

The order of the differential equation

 is
A. 2

B. 1

C. 0

D. not defined

Answer:

It is given that equation is 

We can see that the highest order derivative present in the given differential equation is .

Thus, its order is two.

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Exercise 9.2

Question 1.

In each of the question verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

y = e^x + 1 : y′′ – y′ = 0

Answer:

It is given that y = e^x + 1

Now, differentiating both sides w.r.t. x, we get,

Now, Again, differentiating both sides w.r.t. x, we get,

⇒ y” = e^x

Now, Substituting the values of y’ and y” in the given differential equations, we get,

y” – y’ = e^x - e^x = RHS.

Therefore, the given function is the solution of the corresponding differential equation.

Question 2.

In each of the question verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

y = x² + 2x + C : y′ – 2x – 2 = 0

Answer:

It is given that y = x² + 2x + C

Now, differentiating both sides w.r.t. x, we get,

⇒ y’ = 2x + 2

Now, Substituting the values of y’ in the given differential equations, we get,

y’ – 2x -2 = 2x + 2 – 2x - 2 = 0 = RHS.

Therefore, the given function is the solution of the corresponding differential equation.

Question 3.

In each of the question verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

y = cos x + C : y′ + sin x = 0

Answer:

It is given that y = cosx + C

Now, differentiating both sides w.r.t. x, we get,

⇒ y’ = -sinx

Now, Substituting the values of y’ in the given differential equations, we get,

y’ + sinx = -sinx + sinx = 0 = RHS.

Therefore, the given function is the solution of the corresponding differential equation.

Question 4.

In each of the question verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

Answer:

It is given that y = 

Now, differentiating both sides w.r.t. x, we get,

Therefore, LHS = RHS

Therefore, the given function is the solution of the corresponding differential equation.

Question 5.

In each of the question verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

y = Ax : xy′ = y (x ≠ 0)

Answer:

It is given that y = Ax

Now, differentiating both sides w.r.t. x, we get,

⇒ y’ = A

Now, Substituting the values of y’ in the given differential equations, we get,

xy’ = x.A = Ax = y = RHS.

Therefore, the given function is the solution of the corresponding differential equation.

Question 6.

In each of the question verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

y = x sin x : xy′ = y + x (x ≠ 0 and x > y or x < – y)

Answer:

It is given that y = xsinx

Now, differentiating both sides w.r.t. x, we get,

⇒ y’ = sinx + xcosx

Now, Substituting the values of y’ in the given differential equations, we get,

LHS = xy’ = x(sinx + xcosx)

= xsinx + x²cosx

= y + 

= RHS

Therefore, the given function is the solution of the corresponding differential equation.

Question 7.

In each of the question verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

xy = log y + C : 

Answer:

It is given that xy = log y + C

Now, differentiating both sides w.r.t. x, we get,

⇒ y + xy’ = 

⇒ y² + xyy’ = y’

⇒ (xy – 1)y’ = -y²

⇒ y’ = 

Thus, LHS = RHS

Therefore, the given function is the solution of the corresponding differential equation.

Question 8.

In each of the question verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

y – cos y = x : (y sin y + cos y + x)y′ = y

Answer:

It is given that y – cosy = x

Now, differentiating both sides w.r.t. x, we get,

⇒ y’(1 + siny) = 1

⇒ y’ = 

Now, Substituting the values of y’ in the given differential equations, we get,

LHS = (ysiny + cosy + x)y’

= y = RHS

Therefore, the given function is the solution of the corresponding differential equation.

Question 9.

In each of the question verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

x + y = tan^–1y : y²y′ + y² + 1 = 0

Answer:

It is given that x + y = tan^-1y

Now, differentiating both sides w.r.t. x, we get,

Now, Substituting the values of y’ in the given differential equations, we get,

LHS = y²y’ + y² + 1 = 

= -1 – y² + y² + 1

= 0 = RHS

Therefore, the given function is the solution of the corresponding differential equation.

Question 10.

In each of the question verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

Answer:

It is given that 

Now, differentiating both sides w.r.t. x, we get,

Now, Substituting the values of y’ in the given differential equations, we get,

LHS = 

= x –x = 0 = RHS.

Therefore, the given function is the solution of the corresponding differential equation.

Question 11.

The number of arbitrary constants in the general solution of a differential equation of fourth order are:
A. 0

B. 2

C. 3

D. 4

Answer:

As we know that the number of constant in the general solution of a differential equation of order n is equal to its order.

Thus, the number of constant in the general equation of fourth order differential equation is four.

Question 12.

The number of arbitrary constants in the particular solution of a differential equation of third order are:
A. 3

B. 2

C. 1

D. 0

Answer:

In a particular solution of a differential equation, there is no arbitrary constant.

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Exercise 9.3

Question 1.

In each of the question, form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

Answer:

The given equation is 

Now, differentiating both sides w.r.t x, we get,

Now, again differentiating both sides, we get,

⇒ y’’ = 0

Therefore, the required differential equation is y’’ = 0.

Question 2.

In each of the question, form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

y² = a(b² – x²)

Answer:

The given equation is y² = a(b² – x²)

Now, differentiating both sides w.r.t x, we get,

⇒ 2yy’ = -2ax

⇒ yy’ = -ax -------(1)

Now, again differentiating both sides, we get,

y’.y’ +yy’’ = -a

⇒ (y’)² + yy” = -a --------(2)

Now, dividing equation (2) by (1), we get,

⇒ xyy” + x(y’)² – yy” = 0

Therefore, the required differential equation is xyy” + x(y’)² – yy” = 0.

Question 3.

In each of the question, form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

y = a e^3x + b e^–2x

Answer:

It is given that y = ae^3x + be^-2x --------(1)

Now, differentiating both side we get,

y’ = 3ae^3x - 2be^-2x --------(2)

Now, again differentiating both sides, we get,

y’’ = 9ae^3x + 4be^-2x -------(3)

Now, let us multiply equation (1) with 2 and then adding it to equation (2), we get,

(2ae^3x + 2be^-2x) + (3ae^3x - 2be^-2x) = 2y – y’

⇒ 5ae^3x = 2y + y’

Now, let us multiply equation (1) with 3 and subtracting equation (2), we get

(3ae^3x + 3be^-2x) - (3ae^3x - 2be^-2x) = 3y – y’

⇒ 5be^-2x = 3y - y’

y” = 9. +4

⇒ y” = 6y + y’

⇒ y” – y’ - 6y = 0

Therefore, the required differential equation is y” – y’ - 6y = 0.

Question 4.

In each of the question, form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

y = e^2x (a + bx)

Answer:

It is given y = e^2x(a + bx) -------(1)

Now, differentiating both side w.r.t. x, we get,

y’ = 2e^2x(a + bx) + e^2x.b ------(2)

Now, let us multiply equation (1) with 2 and then subtracting it to equation (2), we get,

y’ – 2y = e^2x(2a +2bx + b) – e^2x(2a + 2bx)

⇒ y’ – 2y = be^2x ---------(3)

Now, again differentiating both sides w.r.t. x, we get,

y” – 2y’ = 2be^2x ------(4)

Dividing equation (4) by equation (3), we get,

⇒ y” – 2y’ = 2y’ – 4y

⇒ y” – 4y’ – 4y = 0

Therefore, the required differential equation is y” – 4y’ - 4y = 0.

Question 5.

In each of the question, form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

y = e^x (a cos x + b sin x)

Answer:

It is given that y = e^x(acosx + bsinx) ------(1)

Now, differentiating both w.r.t. x, we get,

y’ = e^x(acosx + bsinx) + e^x(-asinx + bcosx)

⇒ y’ = e^x[(a + b)cosx – (a – b)sinx)] ------(2)

Again, differentiating both sides w.r.t. x, we get,

y” = e^x[(a + b)cosx – (a – b)sinx)] + e^x[-(a + b)sinx – (a – b)cosx)]

⇒ y” = e^x[2bcosx – 2asinx]

⇒ y” = 2e^x(bcosx – asinx) ----(3)

Adding equation (1) and (3), we get,

⇒ 2y + y” = 2y’

Therefore, the required differential equation is 2y + y” = 2y’= 0.

Question 6.

Form the differential equation of the family of circles touching the y-axis at origin.

Answer:

The center of the circle touching the y- axis at orgin lies on the x – axis.

Let (a,0) be the centre of the circle.

Thus, it touches the y – axis at orgin, its radius is a.

Now, the equation of the circle with centre (a,0) and radius (a) is

(x –a)² – y² = a²

⇒ x² + y² = 2ax

Now, differentiating both sides w.r.t. x , we get,

2x + 2yy’ = 2a

⇒ x + yy’ = a

Now, on substituting the value of a in the equation, we get,

x² + y² = 2(x + yy’)x

⇒ 2xyy’ + x² = y²

Therefore, the required differential equation is 2xyy’ + x² = y² .

Question 7.

Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.

Answer:

We know that the equation of the parabola having the vertex at origin and the axis along the positive y- axis is

x² = 4ay ------(1)

Now, differentiating equation (1) w.r.t. x, we get,

2x = 4ay’ -----(2)

On dividing equation (2) by equation (1), we get,

⇒ xy’ = 2y

⇒ xy’ – 2y = 0

Therefore, the required differential equation is xy’ – 2y = 0.

Question 8.

Form the differential equation of the family of ellipses having foci on y-axis and centre at origin.

Answer:

We know that the equation of the family of ellipses having foci on y – axis and the centre at origin is

 -------(1)

Now, differentiating equation (1) w.r.t. x, we get,

 ----(2)

Now, again differentiating w.r.t. x, we get,

Let us substitute the value in eq. (2), we get,

⇒ -x(y’)² – xyy” + yy’ = 0

⇒ xyy” + x(y’)² – yy’ = 0

Therefore, the required differential equation is xyy” + x(y’)² – yy’ = 0.

Question 9.

Form the differential equation of the family of hyperbolas having foci on x-axis and centre at origin.

Answer:

We know that the equation of the family of hyperbolas having foci on x – axis and the centre at origin is

 -------(1)

Now, differentiating equation (1) w.r.t. x, we get,

 --------(2)

Now, again differentiating w.r.t. x, we get,

Let us substitute the value in eq. (2), we get,

⇒ x(y’)² + xyy” - yy’ = 0

⇒ xyy” + x(y’)² – yy’ = 0

Therefore, the required differential equation is xyy” + x(y’)² – yy’ = 0.

Question 10.

Form the differential equation of the family of circles having centre on y-axis and radius 3 units.

Answer:

Let the centre of the circle on y – axis be (0,b).

We know that the differential equation of the family of circles with centre at (0, b) and radius 3 is: x² + (y- b)² = 3²

⇒ x² + (y- b)² = 9 ----(1)

Now, differentiating both sides w.r.t. x, we get,

2x + 2(y – b).y’ = 0

⇒ (y – b). y’ = x

⇒ y – b = 

Thus, substituting the value of ( y – b) in equation (1), we get,

⇒ x²((y’)² + 1) = 9(y’)²

⇒ (x² – 9)(y’)² + x² = 0

Therefore, the required differential equation is (x² – 9)(y’)² + x² = 0

Question 11.

Which of the following differential equations has y = c₁e^x + c₂e^–x as the general solution?
A. 

B. 

C. 

D. 

Answer:

It is given that y = c₁e^x + c₂e^-x

Now, differentiating both sides w.r.t. x, we get,

Again, differentiating both sides w.r.t. x, we get,

Therefore, the required differential equation is .

Question 12.

Which of the following differential equations has y = x as one of its particular solution?
A. 

B. 

C. 

D. 

Answer:

It is given that that y = x

Now, differentiating both sides w.r.t. x, we get,

Again, differentiating both sides w.r.t. x, we get,

Now, substitute the value of in the given options, we will see that the differential equation in option (C) is correct.

= -x² + x²

= 0

---

Exercise 9.4

Question 1.

For each of the differential equations in question, find the general solution:

Answer:

Question 2.

For each of the differential equations in question, find the general solution:

Answer:

⇒ 

Question 3.

For each of the differential equations in question, find the general solution:

Answer:

dy = (1 - y) dx

Separating variables

⇒ -log(1-y) = x + logc

⇒ -log(1-y) - logc = x

⇒ log (1-y)c = -x

Or

⟹

Question 4.

For each of the differential equations in question, find the general solution:

sec²x tan y dx + sec²y tan x dy

Answer:

Dividing both sides by (tanx)(tany)

Integrating both sides,

⇒ lettan x = t &tany = u

⇒ log t = -log u + log c

Or,

⇒ log(tanx) = -log(tany) + log c

Or

⇒ (tan x) (tan y) = c

Question 5.

For each of the differential equations in question, find the general solution:

(e^x + e^–x)dy – (e^x – e^–x)dx = 0

Answer:

Integrating both sides,

Let ( = t

Then,

So,

⇒ y = log t

Or,

 + c

Question 6.

For each of the differential equations in question, find the general solution:

Answer:

Separating variables,

Integrating both sides,

Question 7.

For each of the differential equations in question, find the general solution:

y log y dx – x dy = 0

Answer:

y log y dx – x dy = 0

⇒ (y log y) dx = xdy

Separating variables,

Integrating both sides,

⇒ let logy = t

⇒ log x + log c = log t

Or,

⇒ log x + log c = log (log y)

⇒ log cx = log y

Or,

⇒ log y = cx

Question 8.

For each of the differential equations in question, find the general solution:

Answer:

Separating variables,

Or,

Integrating both sides,

Letbe a constant,

Or,

0r,

Question 9.

For each of the differential equations in question, find the general solution:

Answer:

Separating variables,

Integrating both sides,

Now to integrate  we have to multiply it by 1
because,

∵ { - i)

So,

According to i)

Let u be and v be 1

We can take the values of u and v from the formula (I.L.A.T.E)

So,

Or

putting the value of t,

Question 10.

For each of the differential equations in question, find the general solution:

e^xtan y dx + 1(1 – e^x)sec²y dy = 0

Answer:

Separating the variables,

Now Integrating both sides,

Let &

Then,

Then,

Or,

⇒ log t = log u + log c

Substituting the values of t and u on above equation.

Or,

Question 11.

For each of the differential equations in question, find a particular solution satisfying the given condition:

(x³ + x² + x + 1) dy/dx = 2x² + x, y = 1 when x = 0

Answer:

(

Separating variables,

Integrating both sides,

Integrating it partially,

Now comparing the coefficients of 

⇒ A + B = 2

⇒ B + C = 1

⇒ A + C = 0

Solving them we will get the values of A,B,C

Putting the values of A,B,C in i)

So,

Then, 2xdx = dt

or,

Or,

Now, we are given that y = 1 when x = 0

So,

C = 1

Putting the value of c in ii)

Question 12.

For each of the differential equations in question, find a particular solution satisfying the given condition:

x(x² – 1) dy/dx = 1; y = 0 when x = 0

Answer:

Separating variables,

Or,

Integrating both sides,

Now let,

Or

Now comparing the values of A,B,C

A + B + C = 0

B-C = 0

A = -1

Solving these we will get that 

Now putting the values of A,B,C in ii)

Now integrating it,

Now we are given that 

Or,

Now putting the value of 

Then,

Question 13.

For each of the differential equations in question, find a particular solution satisfying the given condition:

cos dy/dx = a; y = 2 when x = 0

Answer:

Separating variables,

Now y = 2 when x = 0

⇒ 2 = 0 + c

⇒ c = 2

Putting the value of ini)

Question 14.

For each of the differential equations in question, find a particular solution satisfying the given condition:

dy/dx = y tan x; y = 1 when x = 0

Answer:

Separating variables,

Integrating both sides,

⇒ log y = -log (cos x) + log c

Or,

⇒ log y = log (sec x) + log c

⇒ log y = log c (sec x)

⇒ y = c (sec x) -i)

Now we are given that y = 1 when x = 0

⇒ 1 = c (sec 0)

⇒ 1 = c × 1

⇒ c = 1

Putting the value of c in i)

⇒ y = sec x

Question 15.

Find the equation of a curve passing through the point (0, 0) and whose differential equation is y′ = e^x sin x

Answer:

To find the equation of a curve that passes through point(0,0) and has differential equation 

So, we need to find the general solution of the given differential equation and the put the given point in to find the value of constant.

Separating variables,

Integrating both sides,

{Using the formula, )

Now let 

integrating for .

Or,

Or,

Now we are given that the curve passes through point(0,0)

Putting the value of C in ii)

So,

So the required equations become,

Question 16.

For the differential equation  find the solution curve passing through the point (1, –1).

Answer:

For this question, we need to find the particular solution at point(1,-1) for the given differential equation.

Given differential equation is

Separating variables,

Or,

Integrating both sides,

Now separating like terms on each side,

Or,

Now we are given that, the curve passes through (1, -1)

⇒ -2-c = log (1)

⇒ c = -2 + 0 (∵ log(1) = 0)

So c = -2

Putting the value of c in

Question 17.

Find the equation of a curve passing through the point (0, –2) given that at any point (x, y) on the curve, the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point.

Answer:

We know that slope of a tangent is =

So we are given that the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point.

Now separating variables,

⇒ ydy = xdx

Integrating both sides,

Now the curve passes through (0, -2).

∴ 4-0 = 2c

⇒ c = 2

Putting the value of c in i)

Question 18.

At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, –3). Find the equation of the curve given that it passes through (–2, 1).

Answer:

_{We know that (x,y) is the point of contact of curve and its tangent.}

slope(m1)for line joining (x,y) and (-4,-3) is

Also we know that slope of tangent of a curve is .

Now, according to the question,

(m2) = 2(m1)

Separating variables,

Integrating both sides,

⇒ log(y + 3) = 2log(x + 4) + log c

Now, this equation passes thorough the point (-2,1).

⇒ 4 = 4c

⇒ c = 1

Substitute the value of c in iii)

Question 19.

The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.

Answer:

Let the rate of change of the volume of the balloon be k. (k is a constant)

Or,

Integrating both sides,

Now, given that

At t = 0, r = 3:

⇒ 4π × 33 = 3(k×0 + c)

⇒ 108π = 3c

⇒ c = 36π

At t = 3, r = 6:

⇒ k = 84π

Substituting the values of k and c in i)

So the radius of balloon after t seconds is 

Question 20.

In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs 100 double itself in 10 years (loge2 = 0.6931).

Answer:

let t be time

p be principal

r be rate of interest

according the information principal increases at the rate of r% per year.

Separating variables,

Integrating both sides,

Given that t = 0, p = 100.

Now, if t = 10, then p = 2×100 = 200

So,

So r is 6.93%.

Question 21.

In a bank, principal increases continuously at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years (e^0.5 = 1.648).

Answer:

Let p and t be principal and time respectively.

Given that principal increases continuously at rate of 5% per year.

Separating variables,

Integrating both sides,

When t = 0, p = 1000

At t = 10

So after 10 years the total amount would be Rs.1648

Question 22.

In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?

Answer:

let y be the number of bacteria at any instant t.

Given that the rate of growth of bacteria is proportional to the number present

Separating variables,

Integrating both sides,

⇒ log y = kt + c -i)

Let y’ be the number of bacteria at t = 0.

⇒ log y’ = c

Substituting the value of c in 

⇒ log y = kt + log y’

⇒ log y- log y’ = kt

Also, given that number of bacteria increases by 10% in 2 hours.

Substituting this value in 

So,  becomes

Now, let the time when number of bacteria increase from 100000 to 200000 be t’.

So from 

So bacteria increases from 100000 to 200000 in  hours.

Question 23.

The general solution of the differential equation is
A. e^x + e^–y = C

B. e^x + e^y = C

C. e^–x + e^y = C

D. e^–x + e^–y = C

Answer:

separating variables

Integrating both sides

Or,

 (c is a constant)

So the correct option is A.

---

Exercise 9.5

Question 1.

In each of the question, show that the given differential equation is homogeneous and solve each of them.

(x² + xy)dy = (x² + y²)dx

Answer:

Here, putting x = kx and y = ky

= k⁰.f(x,y)

Therefore, the given differential equation is homogeneous.

(x² + xy)dy = (x² + y²)dx

To solve it we make the substitution.

y = vx

Differentiating eq. with respect to x, we get

Integrating on both side,

- v - 2log|1 - v| = log|x| + logc

The required solution of the differential equation.

Question 2.

In each of the question, show that the given differential equation is homogeneous and solve each of them.

Answer:

Here, putting x = kx and y = ky

= k⁰.f(x,y)

Therefore, the given differential equation is homogeneous.

To solve it we make the substitution.

y = vx

Differentiating eq. with respect to x, we get

v = logx + C

y = xlogx + Cx

The required solution of the differential equation.

Question 3.

In each of the question, show that the given differential equation is homogeneous and solve each of them.

(x – y)dy – (x + y)dx = 0

Answer:

(x - y)dy = (x + y)dx

Here, putting x = kx and y = ky

= k⁰.f(x,y)

Therefore, the given differential equation is homogeneous.

(x - y)dy – (x + y)dx = 0

To make it we make the substitution.

y = vx

Differentiating eq. with respect to x, we get

Integrating both sides we get,

2vdv = dt

The required solution of the differential equation.

Question 4.

In each of the question, show that the given differential equation is homogeneous and solve each of them.

(x² – y²)dx + 2xy dy = 0

Answer:

Here, putting x = kx and y = ky

= k⁰.f(x,y)

Therefore, the given differential equation is homogeneous.

To solve it we make the substitution.

y = vx

Differentiating eq. with respect to x, we get

Integrating both sides, we get

Put 1 + v² = t

2vdv = dt

log(t)

∴ log(1 + v²) = -logx + logC (∴ From (i) eq.)

The required solution of the differential equation.

Question 5.

In each of the question, show that the given differential equation is homogeneous and solve each of them.

Answer:

Here, putting x = kx and y = ky

= k⁰.f(x,y)

Therefore, the given differential equation is homogeneous.

To solve it we make the substitution.

y = vx

Differentiating eq. with respect to x, we get

Integrating both sides, we get

The required solution of the differential equation.

Question 6.

In each of the question, show that the given differential equation is homogeneous and solve each of them.

Answer:

Here, putting x = kx and y = ky

= k⁰.f(x,y)

Therefore, the given differential equation is homogeneous.

To solve it we make the substitution.

y = vx

Differentiating eq. with respect to x, we get

Integrating both sides, we get

The required solution of the differential equation.

Question 7.

In each of the question, show that the given differential equation is homogeneous and solve each of them.

Answer:

Here, putting x = kx and y = ky

= k⁰.f(x,y)

Therefore, the given differential equation is homogeneous.

To solve it we make the substitution.

y = vx

Differentiating eq. with respect to x, we get

Integrating both sides, we get

log secv – logv = 2logkx

The required solution of the differential equation.

Question 8.

In each of the question, show that the given differential equation is homogeneous and solve each of them.

Answer:

Here, putting x= kx and y = ky

= k⁰.f(x,y)

Therefore, the given differential equation is homogeneous.

To solve it we make the substitution.

y = vx

Differentiating eq. with respect to x, we get

Integrating both side, we get

log(cosecv – cotv) = -logx + logC

The required solution of the differential equation.

Question 9.

In each of the question, show that the given differential equation is homogeneous and solve each of them.

Answer:

Here, putting x = kx and y = ky

= k⁰.f(x,y)

Therefore, the given differential equation is homogeneous.

To solve it we make the substitution.

y = vx

Differentiating eq. with respect to x, we get

Integrating both sides, we get

Put, logv – 1 = t

logt

log(logv - 1)

∴ log(logv - 1) – log(v) = log(x) + log(c) (From (i) eq.)

The required solution of the differential equation.

Question 10.

In each of the question, show that the given differential equation is homogeneous and solve each of them.

Answer:

Here, putting x = kx and y = ky

= k⁰f(x,y)

Therefore, the given differential equation is homogeneous.

To solve it we make the substitution.

x = vy

Differentiation eq. with respect to x, we get

Integrating both sides, we get

Put e^v + v = t

(e^v + 1)dv = dt

logt

log(e^v + v)

∴ log(e^v + v) = - logy + logC (∴ From (i) eq.)

Multiply by y on both side, we get

ye^x/y + x = C

x + ye^x/y = C

The required solution of the differential equation.

Question 11.

For each of the differential equations in question, find the particular solution satisfying the given condition:

(x + y)dy + (x – y) dx = 0; y = 1 when x = 1

Answer:

(x + y)dy + (x - y)dx = 0

Here, putting x= kx and y = ky

= k⁰.f(x,y)

Therefore, the given differential equation is homogeneous.

(x + y)dy + (x - y)dx = 0

To solve it we make the substitution.

y = vx

Differentiating eq. with respect to x, we get

Integrating both sides, we get

y = 1 when x = 1

The required solution of the differential equation.

Question 12.

For each of the differential equations in question, find the particular solution satisfying the given condition:

x²dy + (xy + y²)dx = 0; y = 1 when x = 1

Answer:

x²dy + (xy + y²)dx = 0

Here, putting x = kx and y = ky

= k⁰.f(x,y)

Therefore, the given differential equation is homogeneous.

x²dy + (xy + y²)dx = 0

To solve it we make the substitution.

y = vx

Differentiating eq. with respect to x, we get

Integrating both sides, we get

y = 1 when x = 1

3x²y = y + 2x

y + 2x = 3x²y

The required solution of the differential equation.

Question 13.

For each of the differential equations in question, find the particular solution satisfying the given condition:

Answer:

Here, putting x = kx and y = ky

= k⁰.f(x,y)

Therefore, the given differential equation is homogeneous.

To solve it we make the substitution.

y = vx

Differentiating eq. with respect to x, we get

Integrating both sides, we get

∫cosec²vdv = - logx – logC

-cot v = - logx – logC

cot v = logx + logC

1 = C

e¹ = C

The required solution of the differential equation.

Question 14.

For each of the differential equations in question, find the particular solution satisfying the given condition:

y = 0 when x = 1

Answer:

Here, putting x= kx and y = ky

= k⁰.f(x,y)

Therefore, the given differential equation is homogeneous.

To solve it we make the substitution.

y = vx

Differentiating eq. with respect to x, we get

Integrating both sides, we get

- cosv = - logx + C

y = 0 when x = 1

- 1 = C

The required solution of the differential equation.

Question 15.

For each of the differential equations in question, find the particular solution satisfying the given condition:

Answer:

Here, putting x = kx and y = ky

= k⁰.f(x,y)

Therefore, the given differential equation is homogeneous.

To solve it we make the substitution.

y = vx

Differentiating eq. with respect to x, we get

Integrating both sides, we get

y = 2 when x = 1

- 1 = C

The required solution of the differential equation.

Question 16.

A homogeneous differential equation of the from  can be solved by making the substitution.
A. y = vx

B. v = yx

C. x = vy

D. x = v

Answer:

Therefore, we shall substitute,

x = vy

Question 17.

A homogeneous differential equation of the from  can be solved by making the substitution.
A. (4x + 6y + 5)dy – (3y + 2x + 4)dx = 0

B. (xy)dx – (x³ + y³) dy = 0

C. (x³ + 2y²)dx + 2xy dy = 0

D. y²dx + (x² – xy – y²)dy = 0

Answer:

(A) (4x + 6y + 5)dy – (3y + 2x + 4)dx = 0

It cannot be homogeneous as we can see that (4x + 6y + 5) is not homogeneous.

(B) (xy)dx – (x³ + y³)dy = 0

It cannot be homogeneous as the xy which multiplies with dx and x³ + y³ which multiplies with dy are not of same degree.

(C) (x³ + 2y²)dx + 2xy dy = 0

Similarly, it cannot be homogeneous as the x³ + 2y² which multiplies with dx and 2xy which multiplies with dy are not of same degree.

(D) y²dx + (x² – xy – y²)dy = 0

Here, putting x = kx and y = ky

= k⁰.f(x,y)

Therefore, the given differential equations is homogeneous.

---

Exercise 9.6

Question 1.

For each of the differential equations given in question, find the general solution:

Answer:

It is given that 

This is equation in the form of  (where, p = 2 and Q = sinx)

Now, I.F. = 

Thus, the solution of the given differential equation is given by the relation:

y(I.F.) = 

 --------(1)

Let I = 

Now, putting the value of I in (1), we get,

Therefore, the required general solution of the given differential equation is

Question 2.

For each of the differential equations given in question, find the general solution:

Answer:

It is given that 

This is equation in the form of  (where, p = 3 and Q = )

Now, I.F. = 

Thus, the solution of the given differential equation is given by the relation:

y(I.F.) = 

⇒ ye^3x = e^x + C

⇒ y = e^-2x + Ce^-3x

Therefore, the required general solution of the given differential equation is y = e^-2x + Ce^-3x

Question 3.

For each of the differential equations given in question, find the general solution:

Answer:

It is given that 

This is equation in the form of  (where, p =  and Q = )

Now, I.F. = 

Thus, the solution of the given differential equation is given by the relation:

y(I.F.) = 

Therefore, the required general solution of the given differential equation is .

Question 4.

For each of the differential equations given in question, find the general solution:

Answer:

It is given that 

This is equation in the form of  (where, p = secx and Q = tanx)

Now, I.F. = 

Thus, the solution of the given differential equation is given by the relation:

y(I.F.) = 

⇒ y(secx + tanx) = secx + tanx – x+ C

Therefore, the required general solution of the given differential equation is

y(secx + tanx) = secx + tanx – x+ C.

Question 5.

For each of the differential equations given in question, find the general solution:

Answer:

It is given that 

This is equation in the form of  (where, p =  and Q =)

Now, I.F. = 

Thus, the solution of the given differential equation is given by the relation:

y(I.F.) = 

 -----------(1)

Now, Let t = tanx

⇒ sec²xdx = dt

Thus, the equation (1) becomes,

⇒ te^tanx = (t – 1)e^t + C

⇒ te^tanx = (tanx – 1)e^tanx + C

⇒ y = (tanx -1) + C e^-tanx

Therefore, the required general solution of the given differential equation is

y = (tanx -1) + C e^-tanx.

Question 6.

For each of the differential equations given in question, find the general solution:

Answer:

It is given that 

This is equation in the form of  (where, p =  and Q =xlogx)

Now, I.F. = 

Thus, the solution of the given differential equation is given by the relation:

y(I.F.) = 

Therefore, the required general solution of the given differential equation

Question 7.

For each of the differential equations given in question, find the general solution:

Answer:

It is given that 

This is equation in the form of  (where, p =  and Q =)

Now, I.F. = 

Thus, the solution of the given differential equation is given by the relation:

y(I.F.) = 

 ------------------(1)

Now, 

Now, substituting the value in (1), we get,

Therefore, the required general solution of the given differential equation is

Question 8.

For each of the differential equations given in question, find the general solution:

(1 + x²)dy + 2xy dx = cot x dx (x ≠ 0)

Answer:

It is given that 

This is equation in the form of  (where, p =  and Q = )

Now, I.F. = 

Thus, the solution of the given differential equation is given by the relation:

y(I.F.) = 

Therefore, the required general solution of the given differential equation is

Question 9.

For each of the differential equations given in question, find the general solution:

Answer:

It is given that 

This is equation in the form of  (where, p =  and Q = 1 )

Now, I.F. = 

Thus, the solution of the given differential equation is given by the relation:

x(I.F.) = 

⇒ y(xsinx) = -xcosx + sinx + C

⇒ 

⇒ 

Therefore, the required general solution of the given differential equation is

 .

Question 10.

For each of the differential equations given in question, find the general solution:

Answer:

It is given that 

This is equation in the form of  (where, p = -1 and Q = y )

Now, I.F. = 

Thus, the solution of the given differential equation is given by the relation:

x(I.F.) = 

=> x = - y – 1 + Ce^y

=> x + y + 1 = Ce^y

Therefore, the required general solution of the given differential equation is

x + y + 1 = Ce^y.

Question 11.

For each of the differential equations given in question, find the general solution:

y dx + (x – y²)dy = 0

Answer:

It is given that 

This is equation in the form of  (where, p =  and Q = y )

Now, I.F. = 

Thus, the solution of the given differential equation is given by the relation:

x(I.F.) = 

Therefore, the required general solution of the given differential equation is .

Question 12.

For each of the differential equations given in question, find the general solution:

Answer:

It is given that 

This is equation in the form of  (where, p =  and Q = 3y )

Now, I.F. = 

Thus, the solution of the given differential equation is given by the relation:

x(I.F.) = 

⇒ x = 3y² + Cy

Therefore, the required general solution of the given differential equation is x = 3y² + Cy.

Question 13.

For each of the differential equations given in question, find a particular solution satisfying the given condition:

Answer:

It is given that 

This is equation in the form of  (where, p = 2tanx and Q =sinx )

Now, I.F. = 

Thus, the solution of the given differential equation is given by the relation:

y(I.F.) = 

----------------(1)

Now, it is given that y = 0 at x = 

⇒ 0 = 2 + C

⇒ C = -2

Now, Substituting the value of C = -2 in (1), we get,

⇒ y = cosx – 2cos²x

Therefore, the required general solution of the given differential equation is

y = cosx – 2cos²x.

Question 14.

For each of the differential equations given in question, find a particular solution satisfying the given condition:

Answer:

It is given that 

This is equation in the form of  (where, p =  and Q = )

Now, I.F. = 

Thus, the solution of the given differential equation is given by the relation:

y(I.F.) = 

 ----------------(1)

Now, it is given that y = 0 at x = 1

0 = tan^-1 1+ C

⇒ C = 

Now, Substituting the value of C =  in (1), we get,

Therefore, the required general solution of the given differential equation is

Question 15.

For each of the differential equations given in question, find a particular solution satisfying the given condition:

Answer:

It is given that 

This is equation in the form of  (where, p = -3cotx and Q = sin2x)

Now, I.F. = 

Thus, the solution of the given differential equation is given by the relation:

y(I.F.) = 

⇒ y cosec³x = 2cosecx + C

⇒ y = 

⇒ y = -2sin²x + Csin³x--------------(1)

Now, it is given that y = 2 when x = 

Thus, we get,

2 = -2 + C

⇒ C = 4

Now, Substituting the value of C = 4 in (1), we get,

y = -2sin²x + 4sin³x

⇒ y = 4sin³x - 2sin²x

Therefore, the required general solution of the given differential equation is

y = 4sin³x - 2sin²x.

Question 16.

Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the point.

Answer:

Let F(x,y) be the curve passing through origin and let (x,y) be a point on the curve.

We know the slope of the tangent to the curve at (x,y) is .

According to the given conditions, we get,

This is equation in the form of  (where, p = -1 and Q = x )

Now, I.F. = 

Thus, the solution of the given differential equation is given by the relation:

y(I.F.) = 

 --------(1)

Now, 

Thus, from equation (1), we get,

⇒ y = -(x+1) + Ce^x

⇒ x + y + 1 = Ce^x -----(2)

Now, it is given that curve passes through origin.

Thus, equation (2) becomes:

1 = C

⇒ C = 1

Substituting C = 1 in equation (2), we get,

x + y – 1 = e^x

Therefore, the required general solution of the given differential equation is

x + y -1 = e^x

Question 17.

Find the equation of a curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.

Answer:

Let F(x,y) be the curve and let (x,y) be a point on the curve.

We know the slope of the tangent to the curve at (x,y) is .

According to the given conditions, we get,

This is equation in the form of  (where, p = -1 and Q = x - 5)

Now, I.F. = 

Thus, the solution of the given differential equation is given by the relation:

y(I.F.) = 

------------(1)

Now, 

Thus, from equation (1), we get,

⇒ y = 4 – x + Ce^x

⇒ x + y – 4 = Ce^x

Now, it is given that curve passes through (0,2).

Thus, equation (2) becomes:

0 + 2 – 4 = C e⁰

⇒ - 2 = C

⇒ C = -2

Substituting C = -2 in equation (2), we get,

x + y – 4 =-2e^x

⇒ y = 4 – x – 2e^x

Therefore, the required general solution of the given differential equation is

y = 4 – x – 2e^x

Question 18.

The Integrating Factor of the differential equation  is
A. e^–x

B. e^–y

C. 1/x

D. x

Answer:

It is given that 

This is equation in the form of  (where, p =  and Q =2x)

Now, I.F. = 

Question 19.

The Integrating Factor of the differential equation

 is
A. 

B. 

C. 

D. 

Answer:

It is given that 

This is equation in the form of  (where, p =  and Q = )

Now, I.F. = 

---

Miscellaneous Exercise

Question 1.

For each of the differential equations given below, indicate its order and degree (if defined).

Answer:

It is given that equation is 

We can see that the highest order derivative present in the differential is .

Thus, its order is two. It is polynomial equation in . The highest power raised to  is 1.

Therefore, its degree is one.

Question 2.

For each of the differential equations given below, indicate its order and degree (if defined).

Answer:

It is given that equation is 

We can see that the highest order derivative present in the differential is.

Thus, its order is one. It is polynomial equation in . The highest power raised to  is 3.

Therefore, its degree is three.

Question 3.

For each of the differential equations given below, indicate its order and degree (if defined).

Answer:

It is given that equation is 

We can see that the highest order derivative present in the differential is .

Thus, its order is four. The given differential equation is not a polynomial equation.

Therefore, its degree is not defined.

Question 4.

For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

xy = a e^x + b e^–x + x² : 

Answer:

It is given that xy = a e^x + b e^–x + x²

Now, differentiating both sides w.r.t. x, we get,

Now, Again differentiating both sides w.r.t. x, we get,

Now, Substituting the values of ’ and  in the given differential equations, we get,

LHS = 

= x(ae^x +be^-x + 2) + 2(ae^x - be^-x + 2) –x(ae^x +be^-x + x²) + x² – 2

= (axe^x +bxe^-x + 2x) + 2(ae^x - be^-x + 2) –x(ae^x +be^-x + x²) + x² – 2

= 2ae^x -2be^-x +x² + 6x -2

≠ 0

⇒ LHS ≠ RHS.

Therefore, the given function is not the solution of the corresponding differential equation.

Question 5.

For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

y = e^x (a cos x + b sin x) : 

Answer:

It is given that y = e^x(acosx + bsinx) = ae^xcosx + be^xsinx

Now, differentiating both sides w.r.t. x, we get,

Now, again differentiating both sides w.r.t. x, we get,

Now, Substituting the values of ’ and  in the given differential equations, we get,

LHS = 

=2e^x(bcosx – asinx) -2e^x[(a + b)cosx + (b –a ) sinx] + 2e^x(acosx + bsinx)

=e^x[(2bcosx – 2asinx) - (2acosx + 2bcosx) - (2bsinx – 2asinx) + (2acosx + 2bsinx)]

= e^x[(2b – 2a – 2b + 2a)cosx] + e^x[(-2a – 2b + 2a + 2bsinx]

= 0 = RHS.

Therefore, the given function is the solution of the corresponding differential equation.

Question 6.

For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

y = x sin 3x : 

Answer:

It is given that y = xsin3x

Now, differentiating both sides w.r.t. x , we get,

Now, again differentiating both sides w.r.t. x, we get,

Now, substituting the value of  in the LHS of the given differential equation, we get,

= (6.cos3x – 9xsin3x) + 9xsin3x – 6cos3x

= 0 = RHS

Therefore, the given function is the solution of the corresponding differential equation.

Question 7.

For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

x² = 2y² log y : 

Answer:

It is given that x² = 2y² log y

Now, differentiating both sides w.r.t. x, we get,

2x = 2.

Now, substituting the value of  in the LHS of the given differential equation, we get,

= xy –xy

= 0

Therefore, the given function is the solution of the corresponding differential equation.

Question 8.

Form the differential equation representing the family of curves given by (x – a)² + 2y² = a², where a is an arbitrary constant.

Answer:

It is given that (x – a)² + 2y² = a²

⇒ x2 + a2 – 2ax + 2y2 = a2

⇒ 2y2 = 2ax – x2 ---------(1)

Now, differentiating both sides w.r.t. x, we get,

- ---------(2)

So, equation (1), we get,

2ax = 2y² + x²

On substituting this value in equation (3), we get,

Therefore, the differential equation of the family of curves is given as .

Question 9.

Prove that x² – y² = c (x² + y²)² is the general solution of differential equation (x³–3xy²) dx = (y³–3x²y)dy, where c is a parameter.

Answer:

It is given that (x³–3xy²) dx = (y³–3x²y)dy

- --------(1)

Now, let us take y = vx

Now, substituting the values of y and  in equation (1), we get,

On integrating both sides we get,

 --------(2)

Now, 

 ---------(3)

Let 

⇒ 

⇒ 

⇒ 

Now, 

Let v² = p

Now, substituting the values of I₁ and I₂ in equation (3), we get,

Thus, equation (2), becomes,

⇒ (x² – y²)² = C’⁴(x² + y² )⁴

⇒ (x² – y²) = C’²(x² + y² )

⇒ (x² – y²) = C(x² + y² ), where C = C’²

Therefore, the result is proved.

Question 10.

Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes.

Answer:

We know that the equation of a circle in the first quadrant with centre (a, a) and radius a which touches the coordinate axes is :

(x -a)² + (y –a)² = a² -----------(1)

Now differentiating above equation w.r.t. x, we get,

2(x-a) + 2(y-a) = 0

⇒ (x – a) + (y – a)y’ = 0

⇒ x – a +yy’ – ay’ = 0

⇒ x + yy’ –a(1+y’) = 0

⇒ a = 

Now, substituting the value of a in equation (1), we get,

⇒ (x - y)².y’² + (x – y)² = (x + yy’)²

⇒ (x – y)²[1 + (y’)²] = (x + yy’)²

Therefore, the required differential equation of the family of circles is

(x – y)²[1 + (y’)²] = (x + yy’)²

Question 11.

Find the general solution of the differential equation 

Answer:

It is given that 

On integrating, we get,

⇒ sin^-1x + sin^-1y = C

Question 12.

Show that the general solution of the differential equation  is given by (x + y + 1) = A (1 – x – y – 2xy), where A is parameter.

Answer:

It is given that 

On integrating both sides, we get,

Let 

Then,

Now, let A = , then, we have,

Question 13.

Find the equation of the curve passing through the point  whose differential equation is sin x cos y dx + cos x sin y dy = 0.

Answer:

It is given that sin x cos y dx + cos x sin y dy = 0

⇒ tanxdx + tanydy = 0

So, on integrating both sides, we get,

log(secx) + log(secy) = logC

⇒ log(secx.secy) = log C

⇒ secx.secy = C

The curve passes through point 

Thus, 1× = C

⇒ C = 

On substituting C =  in equation (1), we get,

secx.secy = 

Therefore, the required equation of the curve is 

Question 14.

Find the particular solution of the differential equation (1 + e^2x) dy + (1 + y²) e^x dx = 0, given that y = 1 when x = 0.

Answer:

It is given that (1 + e^2x) dy + (1 + y²) e^x dx = 0

On integrating both sides, we get,

 ------(1)

Let e^x = t

⇒ e^2x = t²

⇒ e^xdx = dt

Substituting the value in equation (1), we get,

⇒ tan^-1 y + tan^-1 t = C

⇒ tan^-1 y + tan^-1 (e^x) = C -------(2)

Now, y =1 at x = 0

Therefore, equation (2) becomes:

tan^-1 1 + tan^-1 1 = C

Substituting  in (2), we get,

tan^-1 y + tan^-1 (e^x) = 

Question 15.

Solve the differential equation 

Answer:

It is given that 

 ------(1)

Let 

Differentiating it w.r.t. y, we get,

 ------(2)

From equation (1) and equation (2), we get,

⇒ dz = dy

On integrating both sides, we get,

z = y + C

Question 16.

Find a particular solution of the differential equation (x – y) (dx + dy) = dx – dy, given that y = –1, when x = 0. (Hint: put x – y = t)

Answer:

It is given that (x – y)(dx +dy) = dx – dy

⇒ (x –y + 1)dy = (1- x + y)dx

 ----------------(1)

Let x – y = t

Now, let us substitute the value of x-y and  in equation (1), we get,

 -------(2)

On integrating both side, we get,

t + log|t| = 2x + C

⇒ ( x – y ) + log |x – y| = 2x + C

⇒ log|x – y| = x + y + C --------(3)

Now, y = -1 at x = 0

Then, equation (3), we get,

log 1 = 0 -1 + C

⇒ C = 1

Substituting C = 1in equation (3), we get,

log|x – y| = x + y + 1

Therefore, a particular solution of the given differential equation is log|x – y| = x + y + 1.

Question 17.

Solve the differential equation 

Answer:

It is given that 

This is equation in the form of  (where, p =  and Q = )

Now, I.F. = 

Thus, the solution of the given differential equation is given by the relation:

y(I.F.) = 

Question 18.

Find a particular solution of the differential equation (x ≠ 0), given that y = 0 when 

Answer:

It is given that 

This is equation in the form of  (where, p = cotx and Q = 4xcosecx)

Now, I.F. = 

Thus, the solution of the given differential equation is given by the relation:

y(I.F.) = 

 -----------(1)

Now, y = 0 at x = 

Therefore, equation (1), we get,

0 = 

⇒ C = 

Now, substituting C =  in equation (1), we get,

ysinx = 

Therefore, the required particular solution of the given differential equation is

ysinx = 

Question 19.

Find a particular solution of the differential equation , given that y = 0 when x = 0.

Answer:

It is given that 

On integrating both sides, we get,

 ----------------(1)

Let 

⇒ e^ydt = -dt

Substituting value in equation (1), we get,

⇒ -log|t| = log|C(x+1)|

⇒ -log|2 – e^y| = log|C(x + 1)|

 ------------------(2)

Now, at x = 0 and y = 0, equation (2) becomes,

⇒ C = 1

Now, substituting the value of C I equation (2), we get,

Therefore, the required particular solution of the given differential equation is

Question 20.

The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was 20, 000 in 1999 and 25000 in the year 2004, what will be the population of the village in 2009?

Answer:

Let the population at any instant (t) be y.

Now it is given that the rate of increase of population is proportional to the number of inhabitants at any instant.

Now, integrating both sides, we get,

log y = kt + C ------(1)

According to given conditions,

In the year 1999, t = 0 and y = 20000

⇒ log20000 = C ----(2)

Also, in the year 2004, t = 5 and y = 25000

⇒ log 25000 = k.5 + C

⇒ log 25000 = 5k + log 20000

 --------(3)

Also, in the year 2009, t = 10

Now, substituting the values of t, k and c in equation (1), we get

logy = 

⇒ y = 31250

Therefore, the population of the village in 2009 will be 31250.

Question 21.

The general solution of the differential equation  is
A. xy = C

B. x = Cy²

C. y = Cx

D. y = Cx²

Answer:

It is given that 

Integrating both sides, we get,

log|x| - log|y| = log k

⇒ 

⇒ 

⇒ 

⇒ y = Cx where C = 

Question 22.

The general solution of a differential equation of the type  is
A. 

B. 

C. 

D. 

Answer:

The integrating factor of the given differential equation

 is 

Thus, the general solution of the differential equation is given by,

Question 23.

The general solution of the differential equation e^x dy + (y e^x + 2x) dx = 0 is
A. x ey + x² = C

B. x ey + y² = C

C. y ex + x² = C

D. y ey + x² = C

Answer:

It is given that e^xdy + (ye^x + 2x) dx = 0

This is equation in the form of  (where, p = 1 and Q = -2xe^-x)

Now, I.F. = 

Thus, the solution of the given differential equation is given by the relation:

y(I.F.) = 

⇒ ye^x = -x² + C

⇒ ye^x + x² = C