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Class 12 NCERT Solutions- Mathematics Part II – Chapter 9 Differential Equations-Exercise -9.3
• Last Updated : 05 Mar, 2021

### Question 1.

Solution:

Given:

We can also write

bx + ay = ab

On differentiating we get

b + ay’ = 0

y’ = -b/a

Again differentiating we get

y” = 0

### Question 2.

Solution:

Given:

On differentiating we get

2y.y’=-2ax

Again differentiating we get

xyy” + x(y’)2 – yy’ = 0

### Question 3. y = ae3x+be-2x

Solution:

y = ae3x + be-2x         -(1)

On differentiating we get

y’=3ae3x-2be-2x         -(2)

Again differentiating we get

y”=9ae3x+4be2x

Now on multiply eq(1) by 6

6y = 6ae3x + 6be-2x

And add with eq(2)

6y + y’ = 6ae3x + 6be-2x + 3ae3x – 3be-2x

6y + y’ = 9ae3x + 4be-2z = y”

y” – y’ – 6y = 0

### Question 4. y = e2x(a + bx)

Solution:

Given: y = e2x(a + bx)         -(1)

On differentiating we get

y’ = e2x(b) + (a + bx).2e2x

y’ = e2x(b + 2a + 2bx)        -(2)

Now on multiply eq(1) by 2

2y = e2x(2a + 2bx)

And add with eq(2)

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

y’ – 2y = be2x        -(3)

Again differentiating we get

y” – 2y’ = 2be2x

Now put the value of be2x from eq(3)

y” – 2y’ = 2(y’ – 2y)

y” – 2y’ = 2y’ – 4y

y” – 2y’ – 2y’ + 4y = 0

y” – 4y’ + 4y = 0

### Question 5.

Solution:

Given: y = ex(a cos x + b sin x)         -(1)

On differentiating we get

y’ = ex[a cos x + b sin x – a sin x + b cos x]

y’ = y + ex[b cos x – a sin x]          -(2)

Again differentiating we get

y’ ‘ =y’ + ex[b cos x – a sin x – b sin x – a cos x]

y” = y’ + ex[b cos x – a sin x] – ex[a cos x + b sin x]

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

y” = y’ + [y’ – y] – y

y” – 2y’ + 2y = 0

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

Solution:

Given that the family of circles touching the y-axis at the origin.

So, the center of the circle is (a, 0) and radius a

Let the equation of a circle is

(x – a)2 + y2 = a2

= x2 + y2 = 2ax         -(1)

On differentiating we get

2x + 2yy’ = 2a

x + yy’ = a

Now substitute the value of a in eq(1), we get

x2 + y2 = 2(x + yy’)x

x2 + y2 = 2x2 + 2xyy’

x2 + y2 – 2x2 – 2xyy’

y2 = x2 + 2xyy’

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

Solution:

Given that the family of parabolas having a vertex at origin and axis along positive y-axis.

So the equation of parabola is:

x2 = 4ay           -(1)

On differentiating we get

2x = 4ay’           -(2)

Now divide eq(2) by (1), we have

2x/ x2 = 4ay’ /4ay

2/x = y’ /y

y’x = 2y

y’x – 2y = 0

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

Solution:

Given that the family of ellipses having facii or y-axis & center at the origin.

So the equation of parabola is

On differentiating we get

Again differentiating we get

xyy” + x(y’)2 – yy’ = 0

### Question 9. Form the differential equation of the family of hyperbolas having foci on the x-axis and center at the origin.

Solution:

Given that the family of hyperbolas having foci on the x-axis and center at the origin.

So the equation of hyperbola is

On differentiating we get

Again differentiating we get

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

Solution:

Given that the of circles having a center on y-axis and radius 3 units.

So the center be (0, k)

General equation of the circle is,

x2 + (y – k)2 = 32

(y – k)2 = 9 – x2

y – k =

k =

On differentiating we get

Squaring on both sides we get

### (D)

Solution:

y = c1ex + c2e-x

On differentiating we get

y’ = c1ex – c2e-x

Again differentiating we get

y” = c1ex + c2e-x

y” = y

y” – y = 0

Hence, the correct option is B

### (D)

Solution:

y = x

On differentiating we get

y’ = 1

Again differentiating we get

y” = 0

Now substitute the value of y, y’ and y” in each option to check for correct option

(A)

= 0 – x2(1) + x.x = x

0 ≠ x

(B)

= 0 + x(1) + x.x = x

= x + x2≠ x

(C)

= 0 – x2(1) + x.x = 0

= 0 = 0

Hence, the correct option is C.

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