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 Free PDF Download

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.

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

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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 = ae^3x + be^-2x

Solution:-

4. y = e^2x (a + bx)

Solution:-

From the question, it is given that y = e^2x (a + b x)  … [we call it equation (i)]

Differentiating both sides with respect to x, we get

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

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

We have

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

y’ – 2y = 2ae^2x + 2e^2xbx + e^2xb – 2ae^2x – 2bxe^2x

y’ – 2y = be^2x … [equation (iii)]

Now, differentiating equation (iii) both sides,

We have

⇒ y’’ – 2y = 2be^2x … [equation (iv)]

Then,

5. y = e^x (a cos x + b sin x)

Solution:

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

… [we call it equation (i)]

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)] … [equation (ii)]

Now, differentiating equation (ii) both sides,

We have

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

On simplifying, we get

⇒ y” = e^x[2bcosx – 2asinx]

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

Now, adding equations (i) and (iii), we get

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

Solution:

By looking at the figure, we can say that the centre of the circle touching the y-axis at the origin lies on the x-axis.

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

Hence, it touches the y-axis at the origin, and 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, 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 a vertex at the origin and axis along the positive y-axis.

Solution:

The parabola having the vertex at the 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 the 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 the x-axis and centre at the origin.

Solution:

⇒ x (y’)² + xyy” – yy’ = 0

⇒ xyy” + x(y’)² – yy’ = 0

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

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

Solution:

Let us assume the centre of the circle on the y-axis to 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 e^x + c² e^-x as the general solution?

Solution:

Explanation:

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

Solution:

Explanation:

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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NCERT Solutions for Class 12 Maths

NCERT Solutions for Class 12

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