Class 12 NCERT Solutions- Mathematics Part II – Chapter 9 Differential Equations-Exercise -9.3

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.[Tex]\frac{x}{a}+\frac{y}{b}=1[/Tex]

Solution:

Given: [Tex]\frac{x}{a}+\frac{y}{b}=1[/Tex]

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. [Tex]y^2=a(b^2-x^2)[/Tex]

Solution:

Given: [Tex]y^2=a(b^2-x^2)[/Tex]

On differentiating we get

2y.y’=-2ax

[Tex]\frac{yy’}{x}=-a[/Tex]

Again differentiating we get

[Tex]\frac{x[yy”+(y’)2]-yy’}{x^2}=0[/Tex]

xyy” + x(y’)² – yy’ = 0   

Question 3. y = ae^3x+be^-2x

Solution:

 y = ae^3x + be^-2x         -(1)

On differentiating we get

y’=3ae^3x-2be^-2x         -(2)

Again differentiating we get

y”=9ae^3x+4be^2x

Now on multiply eq(1) by 6 

6y = 6ae^3x + 6be^-2x

And add with eq(2) 

6y + y’ = 6ae^3x + 6be^-2x + 3ae^3x – 3be^-2x

6y + y’ = 9ae^3x + 4be^-2z = y”

y” – y’ – 6y = 0  

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

Solution:

Given: y = e^2x(a + bx)         -(1)

On differentiating we get

y’ = e^2x(b) + (a + bx).2e^2x

y’ = e^2x(b + 2a + 2bx)        -(2)

Now on multiply eq(1) by 2

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

And add with eq(2) 

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

y’ – 2y = be^2x        -(3)

Again differentiating we get

y” – 2y’ = 2be^2x

Now put the value of be^2x from eq(3)

y” – 2y’ = 2(y’ – 2y)

y” – 2y’ = 2y’ – 4y

y” – 2y’ – 2y’ + 4y = 0

y” – 4y’ + 4y = 0

Question 5. [Tex]y=e^x(a\cos x+b\sin x)[/Tex]

Solution:

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

On differentiating we get

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

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

Again differentiating we get

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

y” = y’ + e^x[b cos x – a sin x] – e^x[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 foci on y-axis & center at the origin.

So the equation of parabola is 

[Tex]\frac{x^2}{b^2}+\frac{y^2}{a^2}=1[/Tex]

On differentiating we get

[Tex]\frac{2x}{b^2}+\frac{2y}{a^2}y’=0[/Tex]

[Tex]y’=\frac{-xb^2}{ya^2}[/Tex]

[Tex]\frac{yy’}{x}=\frac{-b^2}{a^2}[/Tex]

Again differentiating we get

[Tex]\frac{x[yy”+(y’)^2]-yy’}{x^2}=0[/Tex]

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 

[Tex]\frac{x^2}{a^2}+\frac{y^2}{b^2}=1[/Tex]

On differentiating we get

[Tex]\frac{2x}{a^2}+\frac{2y}{b^2}y’=0[/Tex]

[Tex]y’=\frac{b^2}{a^2y}[/Tex]

[Tex]\frac{yy’}{x}=\frac{b^2}{a^2}[/Tex]

Again differentiating we get

[Tex]\frac{x[yy”+(y’)^2]-yy’}{x^2}=0[/Tex]

[Tex]y’\sqrt{9-x^2}-x=0  [/Tex]

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 = [Tex]\sqrt{9-x^2}[/Tex]

k = [Tex]y-\sqrt{9-x^2}[/Tex]

On differentiating we get

[Tex]y’-\frac{1}{2\sqrt{9-x^2}}.(-2x)=0[/Tex]

[Tex]y’+\frac{x}{\sqrt{9-x^2}}=0[/Tex]

Squaring on both sides we get

[Tex](9-x^2)(y’)^2+x^2=0[/Tex]

Question 11. Which of the following differential equations has y = c₁e^x + c₂e^-x as the general solution?

(A) [Tex]\frac{d^2y}{dx^2}+y=0[/Tex]

(B) [Tex]\frac{d^2y}{dx^2}-y=0[/Tex]

(C) [Tex]\frac{d^2y}{dx^2}+1=0[/Tex]

(D) [Tex]\frac{d^2y}{dx^2}-1=0[/Tex]

Solution:

y = c₁e^x + c₂e^-x

On differentiating we get

y’ = c₁e^x – c₂e^-x

Again differentiating we get

y” = c₁e^x + 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) [Tex]\frac{d^2y}{dx^2}-x^2\frac{dy}{dx}+xy=x[/Tex]

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

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

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

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

= 0 – x²(1) + x.x = x

0 ≠ x

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

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

= x + x²≠ x

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

= 0 – x²(1) + x.x = 0

= 0 = 0

Hence, the correct option is C.