Class 12 RD Sharma Solutions - Chapter 22 Differential Equations - Exercise 22.7| Set 3

Solve the differential equations(question 40-48):

Question 40. 2x(dy/dx) = 3y, y(1) = 2

Solution:

We have,

2x(dy/dx) = 3y              

2dy/y = 3dx/x

On integrating both sides,

2∫dy/y = 3∫dx/x

2log(y) = 3log(x) + log(c)

y² = x³c

Put x = 1, y = 2 in above equation

c = 4

y² = 4x³ 

Question 41. xy(dy/dx) = y + 2, y(2) = 0

Solution:

We have,

xy(dy/dx) = y + 2       

ydy/(y + 2) = dx/x

On integrating both sides,

∫ydy/(y + 2) = ∫(dx/x)

∫[1 - \frac{2}{(y+2)}  ]dy = ∫(dx/x)

y - 2log(y + 2) = log(x) + log(c)

Put x = 2, y = 0 in above equation

0 - 2log(2) = log(2) + log(c)

log(c) = -3log(2)

log(c) = log(1/8)

c = (1/8)

y - 2log(y + 2) = log(x/8)

Question 42. (dy/dx) = 2e^xy³, y(0) = 1/2

Solution:

We have,

(dy/dx) = 2e^xy³     

dy/y³ = 2e^xdx

On integrating both sides,

∫dy/y³ = 2∫e^xdx

-(1/2y²) = 2e^x + c

Put x = 0, y = (1/2) in above equation

-(4/2) = 2 + c

c = -4

-(1/2y²) = 2e^x - 4

y²(4e^x - 8) = -1

y²(8 - 4e^x) = 1

Question 43. (dr/dt) = -rt, r(0) = r₀

Solution:

We have,

(dr/dt) = -rt       

dr/r = -tdt

On integrating both sides,

∫dr/r = -∫tdt

log(r) = -t²/2 + c

Put t = 0, r =r₀ in above equation

c = log(r₀)

log(r) = -t2/2 + log(r₀)

log(r/r₀) = -t²/2

(r/r₀) = e^{-\frac{t^2}{2}}

r = r₀e^{-\frac{t^2}{2}}

Question 44. (dy/dx) = ysin2x, y(0) = 1

Solution:

We have,

(dy/dx) = ysin2x      

dy/y = sin2xdx

On integrating both sides,

∫(dy/y) = ∫sin2xdx

log(y) = -(1/2)cos2x + c

Put x = 0, y = 1 in above equation

log|1| = -cos0/2 + c

c = (1/2)

log(y) = (1/2) - (cos2x/2)

log(y) = (1 - cos2x)/2

log(y) = 2sin²x/2

log(y) = sin²x

y = e^{sin^2x}

Question 45(i). (dy/dx) = ytanx, y(0) = 1

Solution:

We have,

(dy/dx) = ytanx        

(dy/y) = tanxdx

On integrating both sides

∫(dy/y) = ∫tanxdx

log(y) = log(secx) + c

Put x = 0, y = 1 in above equation

0 = log(1) + c

c = 0

log(y) = log(secx)

y = secx

Question 45(ii). 2x(dy/dx) = 5y, y(1) = 1

Solution:

We have,

2x(dy/dx) = 5y             

(2dy/y) = 5dx/x

On integrating both sides

2∫(dy/y) = 5∫(dx/x)

2log(y) = 5log(x) + c

Put x = 1, y = 1 in above equation

2log(1) = 5log(1) + c

c = 0

2log(y) = 5log(x)

y² = x⁵ 

y = |x|^(5/2)

Question 45(iii). (dy/dx) = 2e^2xy², y(0) = -1

Solution:

We have,

(dy/dx) = 2e^2xy²          

(dy/y2) = 2e^2xdx

On integrating both sides

∫(dy/y²) = 2∫e^2xdx

-(1/y) = 2e^2x/2 + c

-(1/y) = e^2x + c

Put x = 0, y = -1 in above equation

1 = e⁰ + c

c = 0

-(1/y) = e^2x

y = -e^-2x

Question 45(iv). cosy(dy/dx) = e^x, y(0) = π/2

Solution:

We have,

cosy(dy/dx) = e^x         

cosydy = e^xdx

On integrating both sides

∫cosydy = ∫e^xdx

siny = e^x + c

Put x = 1, y = π/2 in above equation

sin(π/2) = e⁰ + c

1 = 1 + c

c = 0

siny = e^x

y = sin^-1(e^x)

Question 45(v). (dy/dx) = 2xy, y(0) = 1

Solution:

We have,

(dy/dx) = 2xy       

dy/y = 2xdx

On integrating both sides

∫(dy/y) = 2∫xdx

log(y) = x² + c

Put x = 0, y = 1 in above equation

log(1) = 0 + c

c = 0

log(y) = x²

Question 45(vi). (dy/dx) = 1 + x² + y² + x²y², y(0) = 1

Solution:

We have,

(dy/dx) = 1 + x² + y² + x²y²     

(dy/dx) = (1 + x²) + y²(1 + x²)

(dy/dx) = (1 + x²)(1 + y²)

\frac{dy}{(1+y^2)}=(1+x^2)dx

On integrating both sides

∫\frac{dy}{(1+y^2)}=∫(1+x^2)dx

tan^-1y = x + (x³/3) + c

Put x = 0, y = 1 in above equation

tan^-1(1) = 0 + 0 + c

c = π/4

tan^-1y = x + (x³/3) + π/4

Question 45(vii). xy(dy/dx) = (x + 2)(y + 2), y(1) = -1

Solution:

We have,

xy(dy/dx) = (x + 2)(y + 2)        

ydy/(y + 2) = (x + 2)dx/x

On integrating both sides

∫ydy/(y + 2) = ∫(x + 2)dx/x

∫[1 - 2/(y + 2)]dy = ∫dx + 2∫(dx/x)

y - 2log(y + 2) = x + 2log(x) + c

Put x = 1, y = -1 in above equation

-1 - 2log(-1 + 2) = 1 + 2log(1) + c

c = -2

y - 2log(y + 2) = x + 2log(x) - 2

Question 45(viii). (dy/dx) = 1 + x + y² + xy², y(0) = 0

Solution:

We have,

(dy/dx) = 1 + x + y² + xy²    

(dy/dx) = (1 + x) + y²(1 + x)

(dy/dx) = (1 + x)(1 + y²)

On integrating both sides

∫\frac{dy}{(1+y^2)}=∫(1+x)dx

tan^-1(y) = x + (x²/2) + c

Put x = 0, y = 0 in above equation

tan^-1(0) = 0 + 0 + c

c = 0

tan^-1(y) = x + (x²/2)

y = tan(x + x²/2)

Question 45(ix). 2(y + 3) - xy(dy/dx) = 0, y(1) = -2

Solution:

We have,

2(y + 3) - xy(dy/dx) = 0   

xy(dy/dx) = 2(y + 3)

ydy/(y + 3) = 2(dx/x)

On integrating both sides

∫[ydy/(y + 3)] = 2∫(dx/x)

∫[1 - 3/(y + 3)]dy = 2∫(dx/x)

y - 3log(y + 3) = 2log(x) + c

Put x = 1, y = -2 in above equation

-2 - 3log(-2 + 3) = 2log(1) + c

c = -2

y - 3log(y + 3) = 2log(x) - 2

y + 2 = log(x)²log(y + 3)³

e^(y+2) = x²(y + 3)³

Question 46. x(dy/dx) + coty = 0, y = π/4 at x = √2

Solution:

We have,

x(dy/dx) + coty = 0                       

x(dy/dx) = -coty

dy/coty = -dx/x

On integrating both sides

∫dy/coty = -∫dx/x

∫tanydy = -∫(dx/x)

log(secy) = -log(x) + c

log(xsecy) = c

Put x = √2, y = π/4 in above equation

log|√2.√2| = c

c = log(2)

log(xsecy) = log(2)

x/cosy = 2

x = 2cosy

Question 47. (1 + x²)(dy/dx) + (1 + y²) = 0, y = 1 at x = 0

Solution:

We have,

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

(1 + x²)(dy/dx) = -(1 + y²)

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

On integrating both sides

∫\frac{dy}{(1+y^2)}=-∫\frac{dx}{(1+x^2)}

tan^-1y = -tan^-1x + c

Put x = 0, y = 1 in above equation

tan^-1(1) = tan^-1(0) + c

c = π/4

tan^-1y = π/4 - tan^-1x

y = tan(π/4 - tan^-1x)

y=\frac{tan\frac{π}{4}-tan(tan^{-1}x)}{1+tan\frac{π}{4}tan(tan^{-1}x)}

y = (1 - x)/(1 + x)

y + yx = 1 - x

x + y = 1 - xy

Question 48. (dy/dx) = 2x(logx + 1)/(siny + ycosy), y = 0 at x = 1  

Solution:

We have,

(dy/dx) = 2x(logx + 1)/(siny + ycosy)                         

(siny + ycosy)dy = 2x(logx + 1)dx

On integrating both sides

∫sinydy + ∫ycosydy = 2∫xlogxdx + 2∫xdx

-cosy + y∫cosydy - ∫[(dy/dy)∫cosydy]dy = 2logx∫xdx - 2∫[(\frac{d(logx)}{dx}  ∫xdx]dx + x² + c

-cosy + ysiny - ∫sinydy = x²logx - ∫xdx + x² + c

-cosy + ysiny + cosy = x²logx - x²/2 + x² + c

Put x = 1, y = 0 in above equation

-1 + 0 + 1 = 0 - (1/2) + 1 + c

c = -(1/2)

ysiny = x²logx + x²/2 - (1/2)

2ysiny = 2x²logx + x² - 1

Question 49. Find the particular solution of the differential equation e^(dy/dx) = x + 1, given that y(0) = 3 when x = 0.

Solution:

We have,

e^(dy/dx) = x + 1               

Taking log both sides,

(dy/dx) = log(x + 1)

dy = log(x + 1)dx

On integrating both sides

∫dy = ∫log(x + 1)dx

y = log(x + 1)∫dx - ∫[\frac{d(log(x+1))}{dx}  ∫dx]dx

y = xlog(x + 1) - ∫xdx/(x + 1)

y = xlog(x + 1) - ∫[1 - 1/(x + 1)]dx

y = xlog(x + 1) - x + log(x + 1) + c

y = (x + 1)log(x + 1) - x + c

Put x = 0, y = 3 in above equation

3 = 0 - 0 + c 

c = 3

y = (x + 1)log(x + 1) - x + 3

Question 50. Find the solution of the differential equation cosydy + cosxsinydx = 0, given that y = π/2 when x = π/2.

Solution:

We have,

cosydy + cosxsinydx = 0         

cosydy = -cosxsinydx

(cosy/siny)dy = -cosxdx

On integrating both sides

∫cotydy = -∫cosxdx

log(siny) = -sinx + c

Put x = π/2, y = π/2 in above equation

log|sinπ/2| = -sin(π/2) + c

0 = -1 + c

c = 1

log(siny) = 1 - sin(x)

log(siny) + sin(x) = 1

Question 51. Find the particular solution of the differential equation (dy/dx) = -4xy², given that y =1  when x = 0.

Solution:

We have,

(dy/dx) = -4xy²           

(dy/y²) + 4xdx = 0

On integrating both sides

∫(dy/y²) + 4∫xdx = 0

-(1/y) + 2x² = c

Put x = 0, y = 1 in above equation

-1 + 0 = c

c = -1

-(1/y) + 2x² = -1

(1/y) = 2x² + 1

y = 1/(2x² + 1)

Question 52. Find the equation of a curve  passing through the point(0, 0) and whose differential equation is  (dy/dx) = e^xsinx.

Solution:

We have,

(dy/dx) = e^xsinx            

dy = e^xsinxdx

On integrating both sides

∫dy = ∫e^xsinxdx

Let, I = ∫e^xsinxdx

I = e^x∫sinx - ∫[\frac{de^x}{dx}  ∫sinxdx]dx

I = -e^xcosx + ∫e^xcosxdx

I = -e^xcosx + e^x∫cosxdx - ∫[\frac{de^x}{dx}  ∫cosxdx]dx

I = -e^xcosx + e^xsinx - ∫e^xsinxdx

I = -e^xcosx + e^xsinx - I

2I = -e^xcosx + e^xsinx

I = e^x(sinx - cosx)/2

y = e^x(sinx - cosx)/2

Question 53. For the differential equation xy(dy/dx) = (x + 2)(y + 2), find the solution curve passing through the point (1, -1).

Solution:

We have,

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

ydy/(y + 2) = (x + 2)dx/x

On integrating both sides

∫ydy/(y + 2) = ∫(x + 2)dx/x

∫[1 - 2/(y + 2)]dy = ∫dx + 2∫(dx/x)

y - 2log(y + 2) = x + 2log(x) + c

y - x - c = log(x)² + log(y + 2)² 

y - x - c = log|x²(y + 2)²|

Curve is passing through (1, -1)

-1 - 1 - c = log(1)

c = 2

y - x - 2 = log|x²(y + 2)²|

Question 54. The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of the balloon after t seconds.

Solution:

We have,

Let, v be the volume of the sphere, t be the time, r be the radius of sphere & k is a constant

Volume of sphere is given by v = (4/3)πr³ 

According to the question (dv/dt) = k

\frac{d[(4/3)πr^3]}{dt} = k

(4/3)π.3r²(dr/dt) = k

4πr²dr = kdt

On integrating both sides

∫4πr²dr = ∫kdt

4π(r³/3) = kt + c

4πr³ = 3(kt + c)           -(i)

At t = 0, r = 38

4π(3)³ = 3(0 + c)

c = 36π

At t = 3, r = 6 in equation (i)

4π(6)³ = 3(kt + 36π)

864π = 9k + 108π

k = 84π

4πr³ = 3(84πt + 36π)

r³ = 63t + 27

r = (63t + 27)^1/3

Radius of the balloon after t second is (63t + 27)^1/3

Question 55. In a bank principal increases at the rate of r % per year. Find the value of r if Rs 100 double itself in 10 years (log 2 = 0.6931).

Solution:

We have,

Let 'p' and 't' be the principal and time respectively.

Principal increases at the rate of r % per year.

dp/dt = (r/100)p

(dp/p) = (r/100)dt

On integrating both sides

∫(dp/p) = (r/100)∫dt

log(p) = (rt/100) + c          -(i)

At t = 0, p = 100

log(100) = 0 + c

c = log(100)          -(ii)

If t = 10, p = 2 × 100 in equation (i)

log(200) = (10r/100) + log(100)

log(200/100) = (10r/100)

log(2) = (r/10)

0.6931 = (r/10)

r = 6.931

Question 56. In a bank principal increases at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years (e = 1.648).

Solution:

We have,

Let 'p' and 't' be the principal and time respectively.

Principal increases at the rate of 5% per year,

(dp/dt) = (5/100)p         -(i)

(dp/p) = (1/20)dt

On integrating both sides

∫(dp/p) = (1/20)∫dt

log(p) = (t/20) + c          -(ii)

At t = 0, p = 1000

log(1000) = c

log(p) = (t/20) + log(1000)

Putting t = 10 in equation in (i)

log(p/1000) = (10/20)

p = 1000e^0.5 

p = 1000 × 1.648

p = 1648

Question 57. In a culture, the bacteria count is 100000. The number is increased by 10% in 2 hours. In how many hours will the count reach 200000, if the rate of growth of bacteria is proportional to the number present?

Solution:

We have,

Let numbers of bacteria at time 't' be 'x'

The rate of growth of bacteria is proportional to the number present

(dx/dt)∝ x          -(i)

(dx/dt) = kx (where 'k' is proportional constant)

(dx/x) = kdt

On integrating both sides

∫(dx/x) = k∫dt

log(x) = kt + c          -(ii)

At t = 0, x = x₀(x₀ is numbers of bacteria at t = 0)

log(x₀) = 0 + c

c = log(x₀)

On putting the value of c in equation (ii)

log(x) = kt + log(x₀)

log(x/x₀) = kt           -(iii)

The number is increased by 10% in 2 hours.

x = x₀(1 + 10/100)

(x/x₀) = (11/10)

On putting the value of (x/x₀) & t = 2 in equation (iii)

2 × k = log(11/10)

k = (1/2)log(11/10)

Therefore, equation (iii) becomes

log(x/x₀) = (1/2)log(11/10) × t

t=\frac{2log\frac{x}{x0}}{log\frac{11}{10}}

At time t₁ numbers of bacteria becomes 200000 from 100000(i.e, x = 2x₀)

t₁ =\frac{2log\frac{2x0}{x0}}{log\frac{11}{10}}

t₁=\frac{2log2}{log\frac{11}{10}}

Question 58. If y(x) is a solution of the differential equation  [\frac{(2 + sinx)}{(1+y)}](\frac{dy}{dx})=-cosx   , and y(0) = 1, then find the value of y(π/2).

Solution:

We have,

 [\frac{(2 + sinx)}{(1+y)}](\frac{dy}{dx})=-cosx                             (i)

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

On integrating both sides

∫dy/(1 + y) = -∫[(cosx)/(2 + sinx)]dx

log(1 + y) = -log(2 + sinx) + log(c)

log(1 + y) + log(2 + sinx) = log(c)

(1 + y)(2 + sinx) = c

Put at x = 0, y = 1

c = (1 + 1)(2 + 0)

c = 4

(1 + y)(2 + sinx) = 4

(1 + y) = 4/(2 + s inx)

y = 4/(2 + sinx) - 1

We need to find the value of y(π/2)

y = 4/(2 + sinπ/2) - 1

y = (4/3) - 1

y = (1/3)

Summary

Exercise 22.7, Set 3 focuses on solving homogeneous differential equations of the first order. These equations are characterized by the right-hand side being a function of y/x. The problems range from simple homogeneous equations to more complex ones involving quadratic expressions. Students are asked to find both general and particular solutions. The primary method used to solve these equations involves substituting y = vx, where v is a function of x, to transform the equation into a separable form. This exercise helps students recognize homogeneous differential equations, apply the appropriate substitution, and solve the resulting separable equation.