# Class 12 RD Sharma Solutions – Chapter 22 Differential Equations – Exercise 22.2 | Set 1

• Last Updated : 16 Feb, 2022

### 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 = emx by eliminating m.

Solution:

y = emx          -(1)

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

dy/dx = memx          -(2)

From eq(1), we get

y = emx

logy = mx

m = (logy/x)

Now, put the value of m in equation(2), we get

x(dy/dx) = ylogy

### (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 + c         -(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 = a         -(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 = Ae2x + Be-2x where A and B are arbitrary constants.

Solution:

y = Ae2x + Be-2x          -(1)

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

(dy/dx) = 2Ae2x – 2Be-2x          -(2)

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

d2y/dx2 = 4Ae2x + 4Be-2x

d2y/dx2 = 4(Ae2x + 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,

x[

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

Solution:

y2 – 2ay + x2 = a         -(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)

-(2)

Let us considered dy/dx = y’

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

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 = r2 by eliminating a and b.

Solution:

(x – a)2 + (y – b)2 = r         -(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

-(3)

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

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

-(4)

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

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 = r

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

(x – a)2 + (y – b)2 = r         -(1)

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

x2 + (y – b)2 = r         -(2)

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

02 + b2 = r         -(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 = 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 = r         -(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 + y

r = (x2 + y2)(2x)          -(3)

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

0 = 2x2 + 2xy(dy/dx) – x2 – y2

(x2 – y2) + 2xy(dy/dx) = 0

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