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

  • Last Updated : 15 Jul, 2021

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 = 4x

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) = 2exy3, y(0) = 1/2

Solution:

We have,

(dy/dx) = 2exy3     

dy/y3 = 2exdx

On integrating both sides,

∫dy/y3 = 2∫exdx

-(1/2y2) = 2ex + c

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

-(4/2) = 2 + c

c = -4

-(1/2y2) = 2ex – 4

y2(4ex – 8) = -1

y2(8 – 4ex) = 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) = 2e2xy2, y(0) = -1

Solution:

We have,

(dy/dx) = 2e2xy         

(dy/y2) = 2e2xdx

On integrating both sides

∫(dy/y2) = 2∫e2xdx

-(1/y) = 2e2x/2 + c

-(1/y) = e2x + c

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

1 = e0 + c

c = 0

-(1/y) = e2x

y = -e-2x

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

Solution:

We have,

cosy(dy/dx) = e        

cosydy = exdx

On integrating both sides

∫cosydy = ∫exdx

siny = ex + c

Put x = 1, y = π/2 in above equation

sin(π/2) = e0 + c

1 = 1 + c

c = 0

siny = ex

y = sin-1(ex)

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 + xy   

(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) = -4xy          

(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) = exsinx.

Solution:

We have,

(dy/dx) = exsinx            

dy = exsinxdx

On integrating both sides

∫dy = ∫exsinxdx

Let, I = ∫exsinxdx

I = ex∫sinx – ∫[\frac{de^x}{dx}  ∫sinxdx]dx

I = -excosx + ∫excosxdx

I = -excosx + ex∫cosxdx – ∫[\frac{de^x}{dx}  ∫cosxdx]dx

I = -excosx + exsinx – ∫exsinxdx

I = -excosx + exsinx – I

2I = -excosx + exsinx

I = ex(sinx – cosx)/2

y = ex(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 = 1000e0.5 

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)

t=\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)


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