# RD Sharma Class 12 Ex 22.2 Solutions Chapter 22 Differential Equations

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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

Solution:

y= (x – c)3

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

2y(dy/dx) = 3(x – c)2

(x – c)= (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

y= [(2y/3)(dy/dx)]3

y4 = (8y3/27)(dy/dx)3

27y= 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:

y= 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 + 2c+ 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)+ (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 = ax+ 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/dx= 2a          -(3)

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

d3y/dx= 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/dx= 4Ae2x + 4Be-2x

d2y/dx= 4(Ae2x + Be-2x)

d2y/dx= 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/dx= -An2cosnt – Bn2sinnt

d2y/dx= -n2(Acosnt + Bsinnt)

d2y/dx+ n2x = 0

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

Solution:

y= 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:

y– 2ay + x= 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

(x– 2y2)y’– 4xyy’ – x= 0

(x– 2y2)(dy/dx)– 4xy(dy/dx) – x= 0

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

Solution:

(x – a)+ (y – b)= 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)– (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)+ (1 + y’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)+ (y – b)= r

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

(x – a)+ (y – b)= r         -(1)

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

x+ (y – b)= r         -(2)

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

0+ b= r         -(3)

x+ (y – b)= r2

On squaring both side, we have

x+ y– 2yr + r= r2          -(Since  r= b2)

2yr = x+ y2

r = (x+ 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

(x– 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)+ (y – b)= r2

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

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

(x – a)+ y= r         -(1)

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

a+ 0= r2          -(2)

(x – a)+ y= r2

On squaring both side, we get,

x– 2ax + a+ y= r2

x+ y– 2xr + r= r2           -(Since  r= a2)

2xr = x+ y

r = (x+ y2)(2x)          -(3)

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

0 = 2x+ 2xy(dy/dx) – x– y2

(x– 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,

= -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)= 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)= 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)+ y= 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

y= 4x+ 4xy(dy/dx)

y– 4x– 4xy(dy/dx) = 0

### (ii) (2x – a)2 – y2 = a2

Solution:

(2x – a)– y= a2

4x– 4ax + a– y= a

4ax = 4x– y2

a = (4x– y2)/4x

On differentiating w.r.t x,

4x+ y= 2xy(dy/dx)

### (iii) (x – a)2 + 2y2 = a2

Solution:

(x – a)+ 2y= 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)

2y– 4xy(dy/dx) – x= 0

### (i) x2 + y2 = a2

Solution:

x+ y= a2

On differentiating w.r.t x,

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

x + y(dy/dx) = 0

### (ii) x2 – y2 = a2

Solution:

x– y= a2

On differentiating w.r.t x,

2x – 2y(dy/dx) = 0

x – y(dy/dx) = 0

### (iii) y2 = 4ax

Solution:

y= 4ax

(y2/x) = 4a

On differentiating w.r.t x,

2xy(dy/dx) – y= 0

2x(dy/dx) – y = 0

### (iv) x2 + (y – b)2 = 1

Solution:

x+ (y – b)= 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)+ x= (dy/dx)2

x2[(dy/dx)+ 1] = (dy/dx)2

### (v) (x – a)2 – y2 = 1

Solution:

(x – a)– y= 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)– y= 1

y2[(dy/dx)2 – 1] = 1

### (vi)

Solution:

We have,

-(1)

{(bx)– (ay)2} = (ab)2           -(2)

On differentiating w.r.t x,

2xb– 2a2y(dy/dx) = 0

xb– a2y(dy/dx) = 0           -(3)

Again, differentiating w.r.t x,

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

xb– a2y(dy/dx) = 0

### (vii) y2 = 4a(x – b)

Solution:

We have,

y= 4a(x – b)

On differentiating w.r.t x,

2y(dy/dx) = 4a

Again differentiating w.r.t x,

[(dy/dx)+ 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,

x+ y= ax3

a = (x+ y2)/(x3

On differentiating w.r.t x,

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

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

2xy(dy/dx) = (x+ 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) × 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)+ (ay)= (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),

xb+ 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)– 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)+ (y – b)= a2         -(1)

x+ 2ax + a+ y– 2ay + a= 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]+ [yp – y – x – yp]= [x + yp]2

(x + y)2p+ (x + y)= (x + yp)2

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

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