Open In App

Class 12 RD Sharma Solutions – Chapter 22 Differential Equations – Exercise 22.3 | Set 2

Last Updated : 03 Mar, 2021
Improve
Improve
Like Article
Like
Save
Share
Report

Question 11: Show that y=(c-x)/(1+cx) is the solution of the differential equation.

(1+x2)(dy/dx)+(1+y2)=0

Solution:

We have,

 y=(c-x)/(1+cx)                (i)

Differentiating equation (i)  w.r.t x,

\frac{dy}{dx}=\frac{(-1)(1+cx)-(c)(c-x)}{(1+cx)^2}

dy/dx=(-1-cx+cx-c2)/(1+cx)2

dy/dx=-(c2+1)/(1+cx2)2

L.H.S,

(1+x2)(dy/dx)+(1+y2)

=(1+x2)[-(c2+1)/(1+cx2)2]+[1+(c-x)2/(1+cx)2]

=-\frac{(1+c^2)(1+x^2)}{(1+cx)^2}+[\frac{(1+cx)^2+(c-x)^2}{(1+cx)^2}]

Simplify the above equation,

=0/(1+cx)2

=0

So, (1+x2)(dy/dx)+(1+y2)=0

Question 12: Show that y=ex(Acosx+Bsinx) is the solution of the differential equation.

d2y/dx2-2(dy/dx)+2y= 0

Solution:

we have,

y=ex(Acosx+Bsinx)               (i)

Differentiating equation (i)  w.r.t x,

dy/dx=ex(Acosx+Bsinx)+ex(-Asinx+Bcosx)              (ii)

dy/dx=ex[(A+B)cosx-(A-B)sinx]                   (iii)

Again differentiating equation (ii)  w.r.t x,

 d2y/dx2 =ex(Acosx+Bsinx)+ex(-Asinx+Bcosx)+ex(-Asinx+Bcosx)+ex(-Acosx-Bsinx)

d2y/dx2=2ex[Bcosx-Asinx]          (iv)

d2y/dx2=2ex[(A+B)cosx-(A-B)sinx] -2ex(Acosx+Bsinx)

d2y/dx2=2(dy/dx)-2y

d2y/dx2-2(dy/dx)+2y= 0

Question 13: Verify that y=cx+2c2 is a solution of the differential equation.

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

Solution:

we have,

y=cx+2c2                (i) 

Differentiating equation (i)  w.r.t x,

dy/dx=c            (ii)

L.H.S,

2(dy/dx)2+x(dy/dx)-y=2(c)2 +x(c)-cx+2c2

=0

Question 14: Verify that y=-x-1 is a solution of the differential equation. 

(y-x)dy-(y2-x2)dx=0

Solution:

we have,

y=-x-1             (i)

Differentiating equation (i)  w.r.t x,

dy/dx=-1

L.H.S,

=(y-x)dy-(y2-x2)dx

=(y-x)(dy/dx)-(y2-x2)

=(-x-1-x)(-1)-[(-x-1)2-x2]

=(2x+1)-(x2+2x+1-x2)

=(x2-x2+2x-2x-1+1)

=0

Question 15: Verify that y2=4a(x+a) is a solution of the differential equation. 

y[1-(dy/dx)2]=2x(dy/dx)

Solution:

we have,

y2=4a(x+a)             (i)

Differentiating equation (i)  w.r.t x,

2y(dy/dx)=4a

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

L.H.S,

=y[1-(dy/dx)2]

=y[1-(2a/y)2

=y[1-(4a2/y2)] 

=y[(y2-4a2)/y2]

=(4a(x+a)-4a2)/y

=(4ax+4a2-4a2)/y  

=[2x(2a)]/y

=2x(dy/dx)

=R.H.S

Question 16: Verify that y=cetan-1 x is a solution of the differential equation. 

(1+x2)(d2y/dx2)+(2x-1)(dy/dx)=0

Solution:

we have,

y=cetan-1 x                 (i)

Differentiating equation (i)  w.r.t x,

dy/dx=cetan-1 x *(1/1+x2)   

(1+x2)(dy/dx)=y               (ii)

Again differentiating equation (ii)  w.r.t x,

2x(dy/dx)+(1+x2)d2y/dx2=dy/dx

(2x-1)(dy/dx)+(1+x2)d2y/dx2=0

Question 17: Verify that y=em cos-1 x is a solution of the differential equation. 

(1-x2)(d2y/dx2)-x(dy/dx)-m2y=0

Solution:

we have,

y=em cos-1 x                (i)

Differentiating equation (i)  w.r.t x,

dy/dx=\frac{me^mcos^{-1}x}{-\sqrt{1-x^2}}

dy/dx=\frac{-my}{\sqrt{1-x^2}}                  (ii)

Again differentiating equation (ii)  w.r.t x,

\frac{d^2y}{dx^2}=\frac{\sqrt{(1-x^2})(-m\frac{dy}{dx})-(-my)\frac{(-2x)}{2\sqrt{1-x^2}}}{(1-x^2)}

\frac{d^2y}{dx^2}=\frac{(-my)(-m)-x\frac{dy}{dx}}{(1-x^2)}

(1-x2)d2y/dx2=m2y-xdy/dx

(1-x2)d2y/dx2-m2y-xdy/dx=0

 Question 18: Verify that y=log(x+1/√(x2+a2))2 is a solution of the differential equation. 

(a2+x2)d2y/dx2+x(dy/dx)=0

Solution:

we have,

y=log(x+1/√(x2+a2))             (i)

Differentiating equation (i)  w.r.t x,

dy/dx=\frac{1}{(x+\sqrt{a^2+x^2})^2}*2(x+\sqrt{a^2+x^2})*\frac{d}{dx}(x+\sqrt{a^2+x^2}))

dy/dx=\frac{2}{(x+\sqrt{a^2+x^2})}*(1+\frac{2x}{2\sqrt{x^2+a^2}})

dy/dx=\frac{2}{(x+\sqrt{x^2+a^2})}\frac{(x+\sqrt{x^2+a^2})}{2\sqrt{x^2+a^2}}

\sqrt{x^2+a^2}(\frac{dy}{dx})=1                      (ii)

Again differentiating equation (ii)  w.r.t x,

(√x2+a2)d2y/dx2+(1/(2√x2+a2))*(2x)*(dy/dx)=0

(a2+x2)d2y/dx2+x(dy/dx)=0

Question 19: Show that the differential equation of which y=2(x2-1)+ce-x2 is the solution 

dy/dx+2xy=4x3

Solution:

we have,

y=2(x2-1)+ce-x2                (i)

Differentiating equation (i)  w.r.t x,

dy/dx=4x+ce-x2(-2x)

dy/dx=4x-2cxe-x2                    (ii)

L.H.S,

=dy/dx+2xy

=4x-2cxe-x2 -2x(y=2(x2-1)+ce-x2   

=4x-2cxe-x2+4x3-4x+2xce-x2

=0

Question 20: Show that y=e-x+ax+c is the solution of the differential equation.

exd2y/dx2=1

Solution:

We have,

 y=e-x+ax+c                       (i)

Differentiating equation (i)  w.r.t x,

dy/dx=-e-x+a                  (ii)

Again differentiating equation (ii)  w.r.t x,

d2y/dx2=e-1

(1/e-1)d2y/dx2=1

exd2y/dx2=1

Question 21: For each of the following differential equations verify that the accompanying function is a solution in the mentioned domain.

(i) Function, y=ax, Differential equation, x(dy/dx)=y 

Solution:

We have,

y=ax                     (i)

Differentiating equation (i)  w.r.t x,

dy/dx=a                  (ii)

From equation (i) a=(y/x)

Putting the value of a in equation (i)

(dy/dx)=a

(dy/dx)=(y/x)

x(dy/dx)=y 

(ii) Function, y=±√(a2-x2), Differential equation: x+y(dy/dx)=0 

Solution:

we have,

y=±√(a2-x2)                     (i)

Squaring both sides, we have

y2=(a2-x2)

2y(dy/dx)=-2x

x+y(dy/dx)=0 

(iii) Function, y=a/(x+a), Differential equation, y+x(dy/dx)=y2 

Solution:

We have,

 y=a/(x+a)                 (i)

Differentiating equation (i)  w.r.t x,

dy/dx=a(-1)/(x+a)2   

dy/dx=-a/(x+a)2   

L.H.S,

 =y+x(dy/dx)

=a/(x+a)-ax/(x+a)

=(-ax+ax+a2)/(x+a)2

=a2/(x+a)2

y2

(iv) Function, y=ax+b+1/2x, Differential equation, x3d2y/dx2=1

Solution:

We have,

y=ax+b+1/2x                 (i)

Differentiating equation (i)  w.r.t x,

dy/dx=a+1/(-2x2)

dy/dx=a-1/2x2                              (ii)

Again differentiating equation (ii)  w.r.t x,

d2y/dx2=0-(-2)/(2x3)

d2y/dx2=1/x3

x3d2y/dx2=1

(v) Function, y=(1/4)*(x±a)2, Differential equation, y=(dy/dx)2 

Solution:

We have,

y=(1/4)*(x±a)2

Differentiating equation (i)  w.r.t x,

dy/dx=(1/4)*2(x±a)

Squaring both side, we have

(dy/dx)2=(1/4)*(x±a)2

(dy/dx)2=y



Previous Article
Next Article

Similar Reads

Class 12 RD Sharma Solutions - Chapter 22 Differential Equations - Exercise 22.9 | Set 3
Question 27. (x2 - 2xy)dy + (x2 - 3xy + 2y2)dx = 0 Solution: We have, (x2 - 2xy)dy + (x2 - 3xy + 2y2)dx = 0 (dy/dx) = (x2 - 3xy + 2y2)/(2xy - y2) It is a homogeneous equation, So, put y = vx (i) On differentiating both sides w.r.t x, dy/dx = v + x(dv/dx) So, v + x(dv/dx) = (x2 - 3xvx + 2v2x2)/(2xvx - x2) v + x(dv/dx) = (1 - 3v + 2v2)/(2v - 1) x(dv/
14 min read
Class 12 RD Sharma Solutions- Chapter 22 Differential Equations - Exercise 22.1 | Set 1
Determine the order and degree of the following differential equation. State also whether it is linear or non-linear(Question 1-13)Question 1. [Tex]\frac{d^3x}{dt^3}+\frac{d^2x}{dt^2}+(\frac{dx}{dt})^2=e^t[/Tex] Solution: We have, [Tex]\frac{d^3x}{dt^3}+\frac{d^2x}{dt^2}+(\frac{dx}{dt})^2=e^t[/Tex] Order of function: The Highest order of derivative
5 min read
Class 12 RD Sharma Solutions - Chapter 22 Differential Equations - Exercise 22.1 | Set 2
Determine the order and degree of the following differential equation. State also whether it is linear or non-linear(Question 14-26)Question 14. [Tex]\sqrt{1-y^2}dx+\sqrt{1-x^2}dy=0[/Tex] Solution: We have, [Tex]\sqrt{1-y^2}dx+\sqrt{1-x^2}dy=0[/Tex] [Tex]\frac{dy}{dx}+ \sqrt{\frac{1-y^2}{1-x^2}}=0[/Tex] Order of function: As the highest order of de
5 min read
Class 12 RD Sharma Solutions - Chapter 22 Differential Equations - Exercise 22.3 | Set 1
Question 1: Show that y=bex+ce2x is the solution of the differential equation. d2 y/dx2-3(dy/dx)+2y=0 Solution: y=bex+ce2x (i) Differentiating equation (i)w.r.t x, dy/dx=bex +2ce2x dy/dx=bex+2ce2x (ii) Again, differentiating equation (ii)w.r.t x, d2y/dx2 =bex+4ce2x (iii) we have, d2y/dx2 -3(dy/dx)+2y=0 (iv) Putting the values ofd2 y/dx2 anddy/dx in
2 min read
Class 12 RD Sharma Solutions - Chapter 22 Differential Equations - Exercise 22.2 | Set 1
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,
6 min read
Class 12 RD Sharma Solutions - Chapter 22 Differential Equations - Exercise 22.2 | Set 2
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 wit
7 min read
Class 12 RD Sharma Solutions - Chapter 22 Differential Equations - Exercise 22.5 | Set 1
Solve the following differential equations:Question 1. (dy/dx) = x2 + x - (1/x) Solution: We have, (dy/dx) = x2 + x - (1/x) dy = [x2 + x - (1/x)]dx On integrating both sides, we get ∫(dy) = ∫[x2 + x - (1/x)]dx y = (x3/3) + (x2/2) - log(x) + c -(Here, 'c' is integration constant) Question 2. (dy/dx) = x5 + x2 - (2/x) Solution: We have, (dy/dx) = x5
5 min read
Class 12 RD Sharma Solutions - Chapter 22 Differential Equations - Exercise 22.5 | Set 2
Question 14. sin4x(dy/dx) = cosx Solution: We have, sin4x(dy/dx) = cosx dy = (cosx/sin4x)dx Let, sinx = z On differentiating both sides, we get cosx dx = dz dy = (dz/z4) On integrating both sides, we get ∫(dy) = ∫(1/z4)dz y = (1/ -3t3) + c y = -(1/3sin3x) + c y = (-cosec3x/3) + c -(Here, 'c' is integration constant) Question 15. cosx(dy/dx) - cos2x
6 min read
Class 12 RD Sharma Solutions - Chapter 22 Differential Equations - Exercise 22.7 | Set 1
Solve the following differential equations:Question 1. (x - 1)(dy/dx) = 2xy Solution: We have, (x - 1)(dy/dx) = 2xy dy/y = [2x/(x - 1)]dx On integrating both sides, ∫(dy/y) = ∫[2x + (x - 1)]dx log(y) = ∫[2 + 2/(x - 1)]dx log(y) = 2x + 2log(x - 1) + c (Where 'c' is integration constant) Question 2. (x2 + 1)dy = xydx Solution: We have, (x2 + 1)dy = x
7 min read
Class 12 RD Sharma Solutions - Chapter 22 Differential Equations - Exercise 22.7 | Set 2
Solve the following differential equations:Question 21. (1 - x2)dy + xydx = xy2dx Solution: We have, (1 - x2)dy + xydx = xy2dx (1 - x2)dy = xy2dx - xydx (1 - x2)dy = xy(y - 1)dx [Tex]\frac{dy}{y(y-1)}=\frac{xdx}{(1-x^2)}[/Tex] On integrating both sides, [Tex]∫[\frac{1}{y-1}-\frac{1}{y}]dy=\frac{1}{2}∫\frac{2x}{1-x^2}dx[/Tex] log(y - 1) - logy = -(1
7 min read