NCERT Solution Class 12 Chapter 9- Differential Equations Exercise 9.3

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

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Access Answers of Maths 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.

NCERT Solutions for Class 12 Maths Chapter 9 -  Image 30

Solution:-

NCERT Solutions for Class 12 Maths Chapter 9 -  Image 31

2. y2 = a (b2 – x2)

Solution:-

NCERT Solutions for Class 12 Maths Chapter 9 -  Image 32

NCERT Solutions for Class 12 Maths Chapter 9 -  Image 33

3. y = ae3x + be-2x

Solution:-

NCERT Solutions for Class 12 Maths Chapter 9 -  Image 34

NCERT Solutions for Class 12 Maths Chapter 9 -  Image 35

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,

NCERT Solutions for Class 12 Maths Chapter 9 -  Image 36

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,

NCERT Solutions for Class 12 Maths Chapter 9 -  Image 37

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

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 -  Image 38

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)

NCERT Solutions for Class 12 Maths Chapter 9 -  Image 39

NCERT Solutions for Class 12 Maths Chapter 9 -  Image 40

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

Solution:

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NCERT Solutions for Class 12 Maths Chapter 9 -  Image 42

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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:

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NCERT Solutions for Class 12 Maths Chapter 9 -  Image 46

NCERT Solutions for Class 12 Maths Chapter 9 -  Image 47

⇒ 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:

NCERT Solutions for Class 12 Maths Chapter 9 -  Image 48

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

NCERT Solutions for Class 12 Maths Chapter 9 -  Image 49

11. Which of the following differential equations has y = c1 ex + c2 e-x as the general solution?

NCERT Solutions for Class 12 Maths Chapter 9 -  Image 50

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 -  Image 51

Explanation:

NCERT Solutions for Class 12 Maths Chapter 9 -  Image 52

NCERT Solutions for Class 12 Maths Chapter 9 -  Image 53

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

NCERT Solutions for Class 12 Maths Chapter 9 -  Image 54

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 -  Image 55

Explanation:

NCERT Solutions for Class 12 Maths Chapter 9 -  Image 56

= 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

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