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

In each of the exercises 1 to 5, from a differential equation representing the given family of curves by eliminating arbitrary constraints a and b.

Question 1.\frac{x}{a}+\frac{y}{b}=1

Solution:

Given: \frac{x}{a}+\frac{y}{b}=1

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. y^2=a(b^2-x^2)

Solution:

Given: y^2=a(b^2-x^2)

On differentiating we get

2y.y’=-2ax



\frac{yy'}{x}=-a

Again differentiating we get

\frac{x[yy''+(y')2]-yy'}{x^2}=0

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. y=e^x(a\cos x+b\sin x)

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 

\frac{x^2}{b^2}+\frac{y^2}{a^2}=1

On differentiating we get

\frac{2x}{b^2}+\frac{2y}{a^2}y'=0

y'=\frac{-xb^2}{ya^2}

\frac{yy'}{x}=\frac{-b^2}{a^2}

Again differentiating we get

\frac{x[yy''+(y')^2]-yy'}{x^2}=0

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 

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1

On differentiating we get

\frac{2x}{a^2}+\frac{2y}{b^2}y'=0

y'=\frac{b^2}{a^2y}

\frac{yy'}{x}=\frac{b^2}{a^2}

Again differentiating we get

\frac{x[yy''+(y')^2]-yy'}{x^2}=0

y'\sqrt{9-x^2}-x=0  

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 = \sqrt{9-x^2}

k = y-\sqrt{9-x^2}

On differentiating we get

y'-\frac{1}{2\sqrt{9-x^2}}.(-2x)=0

y'+\frac{x}{\sqrt{9-x^2}}=0

Squaring on both sides we get

(9-x^2)(y')^2+x^2=0

Question 11. Which of the following differential equations has y = c1ex + c2e-x as the general solution?

(A) \frac{d^2y}{dx^2}+y=0

(B) \frac{d^2y}{dx^2}-y=0

(C) \frac{d^2y}{dx^2}+1=0

(D) \frac{d^2y}{dx^2}-1=0

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

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

(A) \frac{d^2y}{dx^2}-x^2\frac{dy}{dx}+xy=x

(B) \frac{d^2y}{dx^2}+x\frac{dy}{dx}+xy=x

(C) \frac{d^2y}{dx^2}-x^2\frac{dy}{dx}+xy=0

(D) \frac{d^2y}{dx^2}+x\frac{dy}{dx}+xy=0

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) \frac{d^2y}{dx^2}-x^2\frac{dy}{dx}+xy=x

= 0 – x2(1) + x.x = x

0 ≠ x

(B) \frac{d^2y}{dx^2}+x\frac{dy}{dx}+xy=x

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

= x + x2≠ x

(C) \frac{d^2y}{dx^2}-x^2\frac{dy}{dx}+xy=0

= 0 – x2(1) + x.x = 0

= 0 = 0

Hence, the correct option is C.

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