RD Sharma Class 12 Ex 22.2 Solutions Chapter 22 Differential Equations

Here we provide RD Sharma Class 12 Ex 22.2 Solutions Chapter 22 Differential Equations for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 22.2 Solutions Chapter 22 Differential Equations book pdf download. Now you will get step-by-step solutions to each question.

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)

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment in the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rdsharmasolutions.in to your friends.