# Class 12 RD Sharma Solutions – Chapter 22 Differential Equations – Exercise 22.2 | Set 2

Last Updated : 10 May, 2021

### 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,

= -k(4Ï€ r2)

4Ï€r2(dr/dt) = -k(4Ï€r2)

(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

-(2)

Again, differentiating w.r.t x,

d2y/dx2(y – k) + (dy/dx)(dy/dx) = 0

2a(d2y/dx2) + (dy/dx)3 = 0

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

Solution:

-(1)

On differentiating w.r.t x,

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

On putting the value of y in above equation,

=

(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,

Again, on differentiating w.r.t x,

### (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

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 = a

4ax = 4x2 – y2

a = (4x2 – y2)/4x

On differentiating w.r.t x,

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)

2y2 – 4xy(dy/dx) – x2 = 0

### (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,

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

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

Solution:

We have,

-(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,

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

xb2 – a2y(dy/dx) = 0

### (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,

[(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) Ã— 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,

2x3y(dy/dx) = x4 + 3x2y2

2x3y(dy/dx) = x2(x2 + 3y2)

2xy(dy/dx) = (x2 + 3y2)

### (x) y = eax

Solution:

We have,

y = eax         -(1)

On differentiating w.r.t x,

dy/dx = aeax

dy/dx = ay         -(2)

y = eax

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) Ã— 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,

-(where a > b)

(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,

On putting the value of b2 in eq(2),

xb2 + a2y(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

-(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,

Putting the value of b2 in equation(3),

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

On substituting the value of ‘a’ in eq(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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