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

## NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Exercise 9.3 – CBSE Term II Free PDF Download

The Exercise 9.3 of NCERT Solutions for Class 12 Maths Chapter 9- Differential Equations is based on the following topics:

1. Formation of a Differential Equation whose General Solution is given.
2. Procedure to form a differential equation that will represent a given family of curves

Solving the third exercise will help the students improve their hold on these topics and also to score well when the questions are asked from it.

## Download PDF of NCERT Solutions for Class 12 Maths Chapter 9- Differential Equations Exercise 9.3            ### Access Answers to NCERT Class 12 Maths Chapter 9- Differential Equations Exercise 9.3 Page Number 391

In each of the Exercises 1 to 5, form a differential equation representing the given family of curves by eliminating arbitrary constants a and b. Solution:- 2. y2 = a (b2 – x2)

Solution:-  3. y = ae3x + be-2x

Solution:-  4. y = e2x (a + bx)

Solution:-

From the question it is given that y = e2x (a + b x)  … [we call it as equation (i)]

Differentiating both sides with respect to x, we get,

y’ = 2e2x(a + b x) + e2x × b … [equation (ii)]

Then, multiply equation (i) by 2 and afterwards subtract it to equation (ii),

We have,

y’ – 2y = e2x(2a + 2bx + b) – e2x (2a + 2bx)

y’ – 2y = 2ae2x + 2e2xbx + e2xb – 2ae2x – 2bxe2x

y’ – 2y = be2x … [equation (iii)]

Now, differentiating equation (iii) both sides,

We have,

⇒ y’’ – 2y = 2be2x … [equation (iv)]

Then, 5. y = ex (a cos x + b sin x)

Solution:

From the question it is given that y = ex(a cos x + b sin x)

… [we call it as equation (i)]

Differentiating both sides with respect to x, we get,

⇒y’ = ex(a cos x + b sin x) + ex(-a sin x + b cos x)

⇒ y’ = ex[(a + b)cos x – (a – b) sin x)] … [equation (ii)]

Now, differentiating equation (ii) both sides,

We have,

y” = ex[(a + b) cos x – (a – b)sin x)] + ex[-(a + b)sin x – (a – b) cos x)]

On simplifying, we get,

⇒ y” = ex[2bcosx – 2asinx]

⇒ y” = 2ex(b cos x – a sin x) … [equation (iii)]

Now, adding equation (i) and (iii), we get, 6. Form the differential equation of the family of circles touching the y-axis at origin.

Solution: By looking at the figure we can say that the center of the circle touching the y- axis at origin lies on the x – axis.

Let us assume (p, 0) be the centre of the circle.

Hence, it touches the y – axis at origin, its radius is p.

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

⇒ (x – p)2 + y2 = p2

⇒ x2 + p2 – 2xp + y2 = p2

Transposing p2 and – 2xp to RHS then it becomes – p2 and 2xp

⇒ x2 + y2 = p2 – p2 + 2px

⇒ x2 + y2 = 2px … [equation (i)]

Now, differentiating equation (i) both sides,

We have,

⇒ 2x + 2yy’ = 2p

⇒ x + yy’ = p

Now, on substituting the value of ‘p’ in the equation, we get,

⇒ x2 + y2 = 2(x + yy’)x

⇒ 2xyy’ + x2 = y2

Hence, 2xyy’ + x2 = y2 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 parabola having the vertex at origin and the axis along the positive y- axis is

x2 = 4ay … [equation (i)  8. Form the differential equation of the family of ellipses having foci on y-axis and centre at origin.

Solution:    On simplifying,

⇒ -x (y’)2 – xyy” + yy’ = 0

⇒ xyy” + x (y’)2 – yy’ = 0

Hence, xyy” + x (y’)2 – yy’ = 0 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:   ⇒ x (y’)2 + xyy” – yy’ = 0

⇒ xyy” + x(y’)2 – yy’ = 0

Hence, xyy” + x (y’)2 – yy’ = 0is 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 us assume the centre of the circle on y – axis be (0, a).

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

⇒ x2 + (y- a)2 = 9 … [equation (i)]

Now, differentiating equation (i) both sides with respect to x,

⇒ 2x + 2(y – a) × y’ = 0 … [dividing both side by 2]

⇒ x + (y – a) × y’ = 0

Transposing x to the RHS it becomes – x.

⇒ (y – a) × y’ = x 11. Which of the following differential equations has y = c1 ex + c2 e-x as the general solution? Solution: Explanation:  12. Which of the following differential equations has y = x as one of its particular solution? Solution: Explanation: = 0 – (x2 × 1) + (x × x)

= -x2 + x2

= 0

#### Access Other Exercise Solutions of Class 12 Maths Chapter 9

Exercise 9.1 Solutions 12 Questions

Exercise 9.2 Solutions 12 Questions

Exercise 9.4 Solutions 23 Questions

Exercise 9.5 Solutions 17 Questions

Exercise 9.6 Solutions 19 Questions

Miscellaneous Exercise On Chapter 9 Solutions 18 Questions