RD Sharma Class 12 Ex 22.2 Solutions Chapter 22 Differential Equations

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Textbook	NCERT
Class	Class 12th
Subject	Maths
Chapter	22
Exercise	22.2
Category	RD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 22.2 Solutions Chapter 22 Differential Equations

Question 1. Form the differential equation of the family of curves represented by y²= (x – c)³

Solution:

y²= (x – c)³
On differentiating the given equation w.r.t x,
2y(dy/dx) = 3(x – c)²
(x – c)²= (2y/3)(dy/dx)
(x – c) = [(2y/3)(dy/dx)]^1/2-(1)
On putting the value of (x – c) in equation (1), we get
y² = [(2y/3)(dy/dx)]^3/2
On squaring both side, we get
y⁴= [(2y/3)(dy/dx)]³
y⁴ = (8y³/27)(dy/dx)³
27y⁴= 8y³(dy/dx)³
27y = 8(dy/dx)³

Question 2. Form the differential equation corresponding to y = e^mx by eliminating m.

Solution:

y = e^mx -(1)
On differentiating the given equation w.r.t x,
dy/dx = me^mx -(2)
From eq(1), we get
y = e^mx
logy = mx
m = (logy/x)
Now, put the value of m in equation(2), we get
x(dy/dx) = ylogy

Question 3. Form the differential equations from the following primitives where constants are arbitrary.

(i) y²= 4ax

Solution:

y²= 4ax -(1)
y²/4x = a
On differentiating the given equation w.r.t x,
2y(dy/dx) = 4a -(2)
Now, put the value of a in equation(2), we get
2y(dy/dx) = 4(y²/4x)
2y(dy/dx) = y²/x
2x(dy/dx) = y

(ii) y = cx + 2c²+ c³

Solution:

y = cx + 2c²+ c³ -(1)
On differentiating the given equation w.r.t x,
dy/dx = c -(2)
Now, put the value of c in equation(1), we get
y = x(dy/dx) + 2(dy/dx)²+ (dy/dx)³

(iii) xy = a²

Solution:

xy = a² -(1)
On differentiating the given equation w.r.t x,
x(dy/dx) + y = 0

(iv) y = ax²+ bx + c

Solution:

y = ax²+ bx + c   -(1)
On differentiating the given equation w.r.t x,
dy/dx = 2ax + b   -(2)
Again differentiating the given equation w.r.t x,
d²y/dx²= 2a   -(3)
Again, differentiating the given equation w.r.t x, we get
d³y/dx³= 0

Question 4. Find the differential equation of the family of curves y = Ae^2x+ Be^-2x where A and B are arbitrary constants.

Solution:

y = Ae^2x+ Be^-2x   -(1)
On differentiating the given equation w.r.t x,
(dy/dx) = 2Ae^2x– 2Be^-2x   -(2)
Again, differentiating the given equation w.r.t x,
d²y/dx²= 4Ae^2x+ 4Be^-2x
d²y/dx²= 4(Ae^2x+ Be^-2x)
d²y/dx²= 4y

Question 5. Find the differential equation of the family of curves, x = Aconst + Bsinnt where A and B are arbitrary constants.

Solution:

x = Acosnt + Bsinnt -(1)
On differentiating the given equation w.r.t x,
dy/dx = -Ansinnt + Bncosnt -(2)
Again, differentiating the given equation w.r.t x,
d²y/dx²= -An²cosnt – Bn²sinnt
d²y/dx²= -n²(Acosnt + Bsinnt)
d²y/dx²+ n²x = 0

Question 6. Form the differential equation corresponding to y² = a(b – x²) by eliminating a and b.

Solution:

y²= a(b – x²)

On differentiating the given equation w.r.t x,

2y(dy/dx) = a(0 – 2x)

Again, differentiating the given equation w.r.t x,

$2[y\frac{d^2y}{dx^2}+(\frac{dy}{dx})^2]=-2a$

$[y\frac{d^2y}{dx^2}+(\frac{dy}{dx})^2]=-(\frac{y}{-x}*\frac{dy}{dx})$

x[ $x[y\frac{d^2y}{dx^2}+(\frac{dy}{dx})^2]=y\frac{dy}{dx}$

Question 7. Form the differential equation corresponding to y²– 2ay + x²= a²by eliminating a.

Solution:

y²– 2ay + x²= a² -(1)

On differentiating the given equation w.r.t x,

2y(dy/dx) – 2a(dy/dx) + 2x = 0

2y(dy/dx) + 2x = 2a(dy/dx)

$a=\frac{y\frac{dy}{dx}+x}{\frac{dy}{dx}}$ -(2)

Let us considered dy/dx = y’

On putting the value of ‘a’ in the eq(1), we get

$y^2-2[\frac{y\frac{dy}{dx}+x}{\frac{dy}{dx}}]y+x^2=[\frac{y\frac{dy}{dx}+x}{\frac{dy}{dx}}]^2$

$\frac{y'y^2-2y'y^2-2xy+y'x^2}{y'}=\frac{y'^2y^2+x^2+2xyy'}{y'^2}$

On solving this equation, we get

(x²– 2y²)y’²– 4xyy’ – x²= 0

(x²– 2y²)(dy/dx)²– 4xy(dy/dx) – x²= 0

Question 8. Form the differential equation corresponding to (x – a)²+ (y – b)²= r²by eliminating a and b.

Solution:

(x – a)²+ (y – b)²= r² -(1)

On differentiating the given equation w.r.t x,

2(x – a) + 2(y – b)(dy/dx) = 0

(x – a) + (y – b)(dy/dx) = 0 -(2)

Again, differentiating the given equation w.r.t x,

1 + (y – b)(d²y/dx²) + (dy/dx)(dy/dx) = 0

$(y-b)=-[\frac{(\frac{dy}{dx})^2+1}{\frac{d^2y}{dx^2}}]$ -(3)

On putting the value of (y – b) in eq(2),

$(x-a)-[\frac{(\frac{dy}{dx})^2+1}{\frac{d^2y}{dx^2}}]\frac{dy}{dx}=0$

(x – a)(d²y/dx²) – (dy/dx)³– (dy/dx) = 0

$(x-a)=\frac{(\frac{dy}{dx})^3+(\frac{dy}{dx})}{(\frac{d^2y}{dx^2})}$ -(4)

On putting the value of (x – a) and (y – b) in eq(1)

$[\frac{(\frac{dy}{dx})^3+(\frac{dy}{dx})}{(\frac{d^2y}{dx^2})}]^2+ [\frac{(\frac{dy}{dx})^2+1}{\frac{d^2y}{dx^2}}]^2=r^2$

Put (dy/dx) = y’ and d²y/dx² = y”

y’²(1 + y’²)²+ (1 + y’²)²= r²y”²

Question 9. Find the differential equation of all the circles which pass through the origin and whose centres lie on-axis.

Solution:

Equation of the circle is (x – a)²+ (y – b)²= r²

Here, a and b are the centre of the circle.

(x – a)²+ (y – b)²= r² -(1)

When the centre lies on the y-axis, so a = 0

x²+ (y – b)²= r² -(2)

So, when the circle is passing through origin, so the equation is

0²+ b²= r² -(3)

x²+ (y – b)²= r²

On squaring both side, we have

x²+ y²– 2yr + r²= r² -(Since r²= b²)

2yr = x²+ y²

r = (x²+ y²)(2y)

On differentiating the equation(1) w.r.t x, we get

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

0 = 4xy + 4y²(dy/dx) – 2x²(dy/dx) – 2y²(dy/dx)

0 = y²(dy/dx) – x²(dy/dx) + 2xy

(x²– y²)(dy/dx) = 2xy

Question 10. Find the differential equation of all the circles which pass through the origin and whose centres lie on the x-axis.

Solution:

Equation of the circle is (x – a)²+ (y – b)²= r²

Here, a and b are the centre of the circle.

When the centre lies on the x-axis, so b = 0

(x – a)²+ y²= r² -(1)

When the circle is passing through origin, so the equation is

a²+ 0²= r² -(2)

(x – a)²+ y²= r²

On squaring both side, we get,

x²– 2ax + a²+ y²= r²

x²+ y²– 2xr + r²= r² -(Since r²= a²)

2xr = x²+ y²

r = (x²+ y²)(2x) -(3)

On differentiating the equation w.r.t x, we get

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

0 = 2x²+ 2xy(dy/dx) – x²– y²

(x²– y²) + 2xy(dy/dx) = 0

Question 11. Assume that a raindrop evaporates at a rate proportional to its surface area. Form a differential equation involving the rate of change of the radius of the raindrop.

Solution:

Let us considered ‘r’ be the radius of rain drop, volume of the drop be ‘V’ and area of the drop be ‘A’
(dV/dt) proportional to A
(dV/dt) – kA -(V decreases with increasing in t so negative sing)
Here, k is proportionality constant,
$\frac{d(\frac{4π}{3}r^3)}{dt}$ = -k(4π r²)
4πr²(dr/dt) = -k(4πr²)
(dr/dt) = -k

Question 12. Find the differential equation of all the parabolas with latus rectum 4a’ and whose axes are parallel to the x-axis.

Solution:

Equation of parabola whose area is parallel to x-axis and vertices at (h, k).

(y – k)²= 4a(x – h) -(1)

On differentiating w.r.t x,

2(y – k)(dy/dx) = 4a

(y – k)(dy/dx) = 2a

$(y-k)=\frac{2a}{\frac{dy}{dx}}$ -(2)

Again, differentiating w.r.t x,

d²y/dx²(y – k) + (dy/dx)(dy/dx) = 0

$(\frac{2a}{\frac{dy}{dx}})\frac{d^2y}{dx^2}+(\frac{dy}{dx})^2=0$

2a(d²y/dx²) + (dy/dx)³= 0

Question 13. Show that the differential equation of which $y=2(x^2-1)+ce^{-x^2}$ is a solution, is (dy/dx) + 2xy = 4x³

Solution:

$y=2(x^2-1)+ce^{-x^2}$ -(1)

On differentiating w.r.t x,

$\frac{dy}{dx}=4x-2cxe^{-x^2}$

On adding 2xy in R.H.S and L.H.S,

$\frac{dy}{dx}+2xy=4x-2cxe^{-x^2}+2xy$

On putting the value of y in above equation,

$\frac{dy}{dx}+2xy=4x-2cxe^{-x^2}+2x(2(x^2-1)+ce^{-x^2})$

$4x-2cxe^{-x^2}+4x^3-4x+2xce^{-x^2}$ =

(dy/dx) + 2xy = 4x³

Question 14. From the differential equation having y = (sin^-1x)² + A cos^-1x + B, where A and B are arbitrary constants, as its general solution.

Solution:

y = (sin^-1x)² + A cos^-1x + B

On differentiating w.r.t x,

$\frac{dy}{dx} = 2sin^{-1}x(\frac{1}{\sqrt{1-x^2}})+A(\frac{-1}{\sqrt{1-x^2}})+0$

$\sqrt{1-x^2}\frac{dy}{dx}=2sin^{-1}x - A$

Again, on differentiating w.r.t x,

$\sqrt{1-x^2}\frac{d^2y}{dx^2}+\frac{dy}{dx}\frac{1}{\sqrt{1-x^2}}(-2x)\\ = 2(\frac{1}{\sqrt{1-x^2}})-0\\ =(1-x^2)\frac{d^2y}{dx^2}-2x\frac{dy}{dx}-2=0$

Question 15. Form the differential equation of the family of curves represented by the equation (a being the parameter)

(i) (2x + a)²+ y²= a²

Solution:

(2x + a)²+ y²= a² -(1)

On differentiating w.r.t x,

2(2x + a) + 2y(dy/dx) = 0

(2x + a) + y(dy/dx) = 0

a = -2x – y(dy/dx) -(2)

On putting the value of ‘a’ in eq(1), we have

$(2x-2x-y\frac{dy}{dx})^2+y^2=(-2x-y\frac{dy}{dx})^2$

$(y\frac{dy}{dx})^2+y^2=(4x^2+4xy\frac{dy}{dx}+y^2(\frac{dy}{dx})^2)$

y²= 4x²+ 4xy(dy/dx)

y²– 4x²– 4xy(dy/dx) = 0

(ii) (2x – a)²– y²= a²

Solution:

(2x – a)²– y²= a²

4x²– 4ax + a²– y²= a²

4ax = 4x²– y²

a = (4x²– y²)/4x

On differentiating w.r.t x,

$[\frac{4x(8x-2y\frac{dy}{dx}-4(4x^2-y^2}{(4x)^2}]=0$

$32x^2-8xy\frac{dy}{dx}-16x^2+4y^2=0$

$16x^2+4y^2-8xy\frac{dy}{dx}=0$

4x²+ y²= 2xy(dy/dx)

(iii) (x – a)²+ 2y²= a²

Solution:

(x – a)²+ 2y²= a² -(1)

On differentiating w.r.t x,

2(x – a) + 4y(dy/dx) = 0

(x – a) + 2y(dy/dx) = 0

a = x + 2y(dy/dx) -(2)

On putting the value of a in eq(1)

$(x-x-2y\frac{dy}{dx})^2+2y^2=(x+2y\frac{dy}{dx})^2$

$4y^2(\frac{dy}{dx})^2+2y^2=x^2+4xy(\frac{dy}{dx})+4y^2(\frac{dy}{dx})^2$

2y²– 4xy(dy/dx) – x²= 0

Question 16. Represent the following families of curves by forming the corresponding differential equations (a, b being parameters):

(i) x²+ y²= a²

Solution:

x²+ y²= a²
On differentiating w.r.t x,
2x + 2y(dy/dx) = 0
x + y(dy/dx) = 0

(ii) x²– y²= a²

Solution:

x²– y²= a²
On differentiating w.r.t x,
2x – 2y(dy/dx) = 0
x – y(dy/dx) = 0

(iii) y²= 4ax

Solution:

y²= 4ax

(y2/x) = 4a

On differentiating w.r.t x,

$[\frac{2xy\frac{dy}{dx}-y^2}{x^2}]=0$

2xy(dy/dx) – y²= 0

2x(dy/dx) – y = 0

(iv) x²+ (y – b)²= 1

Solution:

x²+ (y – b)²= 1 -(1)

On differentiating w.r.t x,

2x + 2(y – b)(dy/dx) = 0

$(y-b)=-\frac{x}{\frac{dy}{dx}}$

On putting the value of (y – b) in eq(1)

$x^2+[-\frac{x}{\frac{dy}{dx}}]^2=1$

x²(dy/dx)²+ x²= (dy/dx)²

x²[(dy/dx)²+ 1] = (dy/dx)²

(v) (x – a)²– y²= 1

Solution:

(x – a)²– y²= 1 -(1)
On differentiating w.r.t x,
2(x – a) – 2y(dy/dx) = 0
(x – a) – y(dy/dx) = 0
(x – a) = y(dy/dx)
On putting the value of (y – b) in eq(i), we get
y²(dy/dx)²– y²= 1
y²[(dy/dx)2 – 1] = 1

(vi) $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Solution:

We have,

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ -(1)

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

{(bx)²– (ay)²} = (ab)² -(2)

On differentiating w.r.t x,

2xb²– 2a²y(dy/dx) = 0

xb²– a²y(dy/dx) = 0 -(3)

Again, differentiating w.r.t x,

$b^2-a^2[y(\frac{dy}{dx})(\frac{dy}{dx})+y\frac{d^2y}{dx^2}]$

$b^2=a^2[y(\frac{dy}{dx})(\frac{dy}{dx})+y\frac{d^2y}{dx^2}]$

On putting the value of b² in equation(3), we get

xb²– a²y(dy/dx) = 0

$xa^2[y(\frac{dy}{dx}^2)+y\frac{d^2y}{dx^2}]-a^2y\frac{dy}{dx}=0$

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

$x[y\frac{d^2y}{dx^2}+(\frac{dy}{dx})^2]=y(\frac{dy}{dx})$

(vii) y²= 4a(x – b)

Solution:

We have,

y²= 4a(x – b)

On differentiating w.r.t x,

2y(dy/dx) = 4a

Again differentiating w.r.t x,

$2[(\frac{dy}{dx})(\frac{dy}{dx})+y\frac{d^2y}{dx^2}]=0$

[(dy/dx)²+ y(d²y/dx²)] = 0

(viii) y = ax³

Solution:

We have,
y = ax³ -(1)
On differentiating w.r.t x,
(dy/dx) = 3ax²
From eq(1),
a=(y/x³ -(1)
On putting the value of a in eq(1)
dy/dx = 3(y/x³) × x²
x(dy/dx) = 3y

(ix) x²+ y²= ax³

Solution:

We have,

x²+ y²= ax³

a = (x²+ y²)/(x³)

On differentiating w.r.t x,

$\frac{(2x+2y\frac{dy}{dx})(x^3)-(3x^2)(x^2+y^2)}{(x^3)^2}=0$

$2x^3y\frac{dy}{dx}+2x^4-3x^4-3x^2y^2=0$

2x³y(dy/dx) = x⁴+ 3x²y²

2x³y(dy/dx) = x²(x²+ 3y²)

2xy(dy/dx) = (x²+ 3y²)

(x) y = e^ax

Solution:

We have,
y = e^ax -(1)
On differentiating w.r.t x,
dy/dx = ae^ax
dy/dx = ay -(2)
y = e^ax
On taking log both side, we get
logy = ax
a = (logy/x)
Now, put the value of ‘a’ in eq(2)
(dy/dx) = logy/x) × y
x(dy/dx) = ylogy

Question 17. Form the differential equation representing the family of ellipses having foci on the x-axis and the centre at the origin.

Solution:

We have,

Equation of ellipse having foci on the x-axis,

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ -(where a > b)

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

(bx)²+ (ay)²= (ab)² -(1)

On differentiating above equation w.r.t x,

2b²x + 2a²y(dy/dx) = 0

b²x + a²y(dy/dx) = 0 -(2)

Again, differentiating w.r.t x,

$b^2+a^2[y(\frac{dy}{dx})(\frac{dy}{dx})+y\frac{d^2y}{dx^2}]=0$

$b^2=-a^2[y(\frac{dy}{dx})(\frac{dy}{dx})+y\frac{d^2y}{dx^2}]$

On putting the value of b² in eq(2),

xb²+ a²y(dy/dx) = 0

$-xa^2[y(\frac{dy}{dx}^2)+y\frac{d^2y}{dx^2}]+a^2y\frac{dy}{dx}=0$

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

x[y(d²y/dx²) + (dy/dx)²] = y(dy/dx)

Question 18. Form the differential equation of the family of hyperbolas having foci on the X-axis and centre at the origin

Solution:

We have,

Equation of a hyperbola having a Centre at the origin and foci along x-axis

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ -(1)

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

(bx)²-(ay)²=(ab)² -(2)

On differentiating above equation w.r.t x,

2xb²-2a²y(dy/dx)=0

xb²-a²y(dy/dx)=0 -(3)

Again, differentiating above equation w.r.t x,

$b^2-a^2[y(\frac{dy}{dx})(\frac{dy}{dx})+y\frac{d^2y}{dx^2}]=0$

$b^2=a^2[y(\frac{dy}{dx})(\frac{dy}{dx})+y\frac{d^2y}{dx^2}]$

Putting the value of b² in equation(3),

$xb^2-a^2y\frac{dy}{dx}=0$

$xa^2[y(\frac{dy}{dx}^2)+y\frac{d^2y}{dx^2}]-a^2y\frac{dy}{dx}=0$

xy(d²y/dx²) + x(dy/dx)²– y(dy/dx) = 0

x[y(d²y/dx²) +(dy/dx)²] = y(dy/dx)

This is required differential equation.

Question 19. Form the differential equation of the family of circles in the second quadrant and touching the coordinate axis.

Solution:

We have,

Let (-a, a) be the coordinates of the centre of circle

So, the equation of circle is given by,

(x + a)²+ (y – b)²= a² -(1)

x²+ 2ax + a²+ y²– 2ay + a²= 0 -(2)

On differentiating above equation w.r.t x,

2x + 2a + 2y(dy/dx) – 2a(dy/dx) = 0

x + a + y(dy/dx) – a(dy/dx) = 0

$x+y\frac{dy}{dx}=a(\frac{dy}{dx}-1)$

$a=\frac{x+y\frac{dy}{dx}}{\frac{dy}{dx}-1}$

On substituting the value of ‘a’ in eq(2)

$[x+\frac{x+y\frac{dy}{dx}}{\frac{dy}{dx}-1}]^2+[y-\frac{x+y\frac{dy}{dx}}{\frac{dy}{dx}-1}]^2=[\frac{x+y\frac{dy}{dx}}{\frac{dy}{dx}-1}]^2$

Let, (dy/dx) = p

[xp – x + x + yp]²+ [yp – y – x – yp]²= [x + yp]²

(x + y)²p²+ (x + y)²= (x + yp)²

(x + y)²[p²+ 1] = (x + yp)² -(where (dy/dx) = p)

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RD Sharma Class 12 Ex 22.2 Solutions Chapter 22 Differential Equations

Question 1. Form the differential equation of the family of curves represented by y2 = (x – c)3

Question 2. Form the differential equation corresponding to y = emx by eliminating m.

Question 3. Form the differential equations from the following primitives where constants are arbitrary.

(i) y2 = 4ax

(ii) y = cx + 2c2 + c3

(iii) xy = a2

(iv) y = ax2 + bx + c

Question 4. Find the differential equation of the family of curves y = Ae2x + Be-2x where A and B are arbitrary constants.

Question 5. Find the differential equation of the family of curves, x = Aconst + Bsinnt where A and B are arbitrary constants.

Question 6. Form the differential equation corresponding to y2 = a(b – x2) by eliminating a and b.

Question 7. Form the differential equation corresponding to y2 – 2ay + x2 = a2 by eliminating a.

Question 8. Form the differential equation corresponding to (x – a)2 + (y – b)2 = r2 by eliminating a and b.

Question 9. Find the differential equation of all the circles which pass through the origin and whose centres lie on-axis.

Question 10. Find the differential equation of all the circles which pass through the origin and whose centres lie on the x-axis.

Question 11. Assume that a raindrop evaporates at a rate proportional to its surface area. Form a differential equation involving the rate of change of the radius of the raindrop.

Question 12. Find the differential equation of all the parabolas with latus rectum 4a’ and whose axes are parallel to the x-axis.

Question 13. Show that the differential equation of which is a solution, is (dy/dx) + 2xy = 4x3

Question 14. From the differential equation having y = (sin-1x)2 + A cos-1x + B, where A and B are arbitrary constants, as its general solution.

Question 15. Form the differential equation of the family of curves represented by the equation (a being the parameter)

(i) (2x + a)2 + y2 = a2

(ii) (2x – a)2 – y2 = a2

(iii) (x – a)2 + 2y2 = a2

Question 16. Represent the following families of curves by forming the corresponding differential equations (a, b being parameters):

(i) x2 + y2 = a2

(ii) x2 – y2 = a2

(iii) y2 = 4ax

(iv) x2 + (y – b)2 = 1

(v) (x – a)2 – y2 = 1

(vi)

(vii) y2 = 4a(x – b)

(viii) y = ax3

(ix) x2 + y2 = ax3

(x) y = eax

Question 17. Form the differential equation representing the family of ellipses having foci on the x-axis and the centre at the origin.

Question 18. Form the differential equation of the family of hyperbolas having foci on the X-axis and centre at the origin

Question 19. Form the differential equation of the family of circles in the second quadrant and touching the coordinate axis.

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Question 1. Form the differential equation of the family of curves represented by y²= (x – c)³

Question 2. Form the differential equation corresponding to y = e^mx by eliminating m.

(i) y²= 4ax

(ii) y = cx + 2c²+ c³

(iii) xy = a²

(iv) y = ax²+ bx + c

Question 4. Find the differential equation of the family of curves y = Ae^2x+ Be^-2x where A and B are arbitrary constants.

Question 6. Form the differential equation corresponding to y² = a(b – x²) by eliminating a and b.

Question 7. Form the differential equation corresponding to y²– 2ay + x²= a²by eliminating a.

Question 8. Form the differential equation corresponding to (x – a)²+ (y – b)²= r²by eliminating a and b.

Question 13. Show that the differential equation of which $y=2(x^2-1)+ce^{-x^2}$ is a solution, is (dy/dx) + 2xy = 4x³

Question 14. From the differential equation having y = (sin^-1x)² + A cos^-1x + B, where A and B are arbitrary constants, as its general solution.

(i) (2x + a)²+ y²= a²

(ii) (2x – a)²– y²= a²

(iii) (x – a)²+ 2y²= a²

(i) x²+ y²= a²

(ii) x²– y²= a²

(iii) y²= 4ax

(iv) x²+ (y – b)²= 1

(v) (x – a)²– y²= 1

(vi) $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

(vii) y²= 4a(x – b)

(viii) y = ax³

(ix) x²+ y²= ax³

(x) y = e^ax