# Class 12 Maths NCERT Solutions for Chapter 9 Differential Equations Exercise 9.3

### Differential Equations Exercise 9.3 Solutions

**1. Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.x/a + y/b = 1 **

**Solution**

x/a + y/b = 1

Differentiating both sides of the given equation with respect to x, we get :

1/a + (1/b)(dy/dx) = 0

⇒ 1/a + (1/b)y' = 0

Again , differentiating both sides with respect to x, we get :

0 + (1/b)y" = 0

⇒ (1/b)y" = 0

⇒ y" = 0

Hence, the required differential equation of the given curve is y" = 0.

**2. Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.y ^{2} = a (b^{2} – x^{2})**

**Solution**

y^{2} = a (b^{2} – x^{2})

Differentiating both sides with respect to x, we get:

⇒ 2yy' = -2ax

⇒ yy' = -ax** ...(1)**

Again, differentiating both sides with respect to x, we get :

y'. y' + yy" = -a

⇒ (y')^{2} + yy" = -a **...(2) **

Dividing equation (2) by equation (1), we get :

⇒ xyy" + x(y')^{2} - yy" = 0

This is the required differential equation of the given curve.

**3. Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.y = a e ^{3x} + b e^{–}^{2x}. **

**Solution**

y' = 3ae^{3x} - 2be^{-2x}** ...(2)**

Again, differentiating both sides with respect to x, we get :

Multiplying equation (1) with 2 and then adding it to equation (2) , we get :

Now, multiplying equation (1) with 3 and subtracting equation (2) from it, we get :

Substituting the values of ae^{3x} and be^{-2x} in equation (3), we get :

⇒ y'' = 6y + y'

⇒ y'' - y' - 6y = 0

This is the required differential equation of the given curve.

**4. Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.y = e ^{2x} (a + bx)**

**Solution**

y = e^{2x} (a + bx)

Differentiating both sides with respect to x, we get :

y' = 2e^{2x} (a + bx) + e^{2x}.b

⇒ y' = e^{2x} (2a + 2bx + b) **...(2) **

Multiplying equation (1) with 2 and then subtracting it from equation (2),we get :

Differentiating both sides with respect to x, we get :

y" - 2y' = 2be^{2x} **...(4) **

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

This is the required differential equation of the given curve.

**5. 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)**

**Solution**

y = e^{x} (a cos x + b sin x)

Differentiating both sides with respect to x, we get :

y' = e^{x} (a cos x + b sin x) + e^{x} (- a sin x + b cos x)

= y' = e^{x} [(a + b) cos x - (a - b) sin x] **...(2) **

Again, differentiating with respect to x, we get :

Adding equations (1) and (3) , we get :

⇒ 2y + y" = 2y'

⇒ y" - 2y' + 2y = 0

This is the required differential equation of the given curve.

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

**Solution**

The centre of the circle touching the y - axis at origin lies on the x - axis.

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

Since it touches the y - axis at origin, its radius is a.

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

Differentiating equation (1) with respect to x, we get :

2x + 2yy' = 2a

⇒ x + yy' = a

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

This is the required differential equation.

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

**Solution**

The equation of the parabola having the vertex at origin and the axis along the positive y axis is :

x^{2} = 4ay **...(1)**

Differentiating equation (1) with respect to x, we get :

2x = 4ay' **...(2)**

Dividing equation (2) by equation (1), we get :

This is the required differential equation.

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

**Solution**

The equation of the family of ellipses having foci on the y - axis and the centre at origin is as follows:

x^{2} /b^{2} + y^{2} /a^{2} = 1 .**..(1) **

Differentiating equation (1) with respect to x, we get :

Again, differentiating with respect to x, we get :

Substituting this value in equation (2), we get :

This is the required differential equation.

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

**Solution**

The equation of the family of hyperbolas with the centre at origin and foci along the x - axis is :

x^{2} /a^{2} + y^{2} /b^{2} = 1 **...(1) **

Differentiating both sides of equation (1) with respet to x, we get :

Again, differentiating both sides with respect to x, we get :

Substituting the value of 1/a^{2} in equation (2), we get :

This is the required differential equation.

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

**Solution**

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

The differential equation of the family of circles with centre at (0, b) and radius 3 is as follows :

x^{2} + (y - b)^{2} = 3^{2}

⇒ x^{2} + (y - b)^{2} = 9 **...(1) **

Differentiating equation (1) with respect to x, we get :

2x + 2(y - b). y' = 0

⇒ (y - b). y' = -x

⇒ y - b = -x/y'

Substituting the value of (y - b) in equation (1), we get :

**11. Which of the following differential equations has y = c**

_{1}e^{x}+ c_{2}e^{–x}as the general solution ?**Solution**

y = c

_{1}e

^{x}+ c

_{2}e

^{–x}

**...(1)**

Differentiating with respect to x, we get :

dy/dx = c

_{1}e

^{x}- c

_{2}e

^{–x}

Again, differentiating with respect to x, we get :

This is the required differential equation of the given equation of curve.

Hence, the correct answer is B.

**12. Which of the following differential equation has y = x as one of its particular solution ?**

**Solution**

Differentiating with respect to x, we get :

dy/dx =1

**...(1)**

Again, differentiating with respect to x, we get :

d

^{2}y/dx

^{2}= 0

**...(2)**

Now, on substituting the values of y, d

^{2}y/dx

^{2}, and dy/dx from equation (1) and (2) in each of the given alternatives, we find that only the differential equation given in alternative C is correct.

= -x

^{2}+ x

^{2}

= 0

Hence, the correct answer is C.