# 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)

y

^{2 }= x^{3}cPut x = 1, y = 2 in above equation

c = 4

y

^{2 }= 4x^{3 }

### 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^{x}y^{3}, y(0) = 1/2

**Solution:**

We have,

(dy/dx) = 2e

^{x}y^{3}dy/y

^{3 }= 2e^{x}dxOn integrating both sides,

∫dy/y

^{3 }= 2∫e^{x}dx-(1/2y

^{2}) = 2e^{x }+ cPut x = 0, y = (1/2) in above equation

-(4/2) = 2 + c

c = -4

-(1/2y

^{2}) = 2e^{x }– 4y

^{2}(4e^{x }– 8) = -1y

^{2}(8 – 4e^{x}) = 1

### Question 43. (dr/dt) = -rt, r(0) = r_{0}

**Solution:**

We have,

(dr/dt) = -rt

dr/r = -tdt

On integrating both sides,

∫dr/r = -∫tdt

log(r) = -t

^{2}/2 + cPut t = 0, r =r

_{0}in above equationc = log(r

_{0})log(r) = -t2/2 + log(r

_{0})log(r/r

_{0}) = -t^{2}/2(r/r

_{0}) =r = r

_{0}

### 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) = 2sin

^{2}x/2log(y) = sin

^{2}xy =

### 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)

y

^{2 }= x^{5}y = |x|

^{(5/2)}

### Question 45(iii). (dy/dx) = 2e^{2x}y^{2}, y(0) = -1

**Solution:**

We have,

(dy/dx) = 2e

^{2x}y^{2 }(dy/y2) = 2e

^{2x}dxOn integrating both sides

∫(dy/y

^{2}) = 2∫e^{2x}dx-(1/y) = 2e

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

^{2x }+ cPut x = 0, y = -1 in above equation

1 = e

^{0 }+ cc = 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

^{x}dxOn integrating both sides

∫cosydy = ∫e

^{x}dxsiny = e

^{x }+ cPut x = 1, y = π/2 in above equation

sin(π/2) = e

^{0 }+ c1 = 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) = x

^{2 }+ cPut x = 0, y = 1 in above equation

log(1) = 0 + c

c = 0

log(y) = x

^{2}

### Question 45(vi). (dy/dx) = 1 + x^{2 }+ y^{2 }+ x^{2}y^{2}, y(0) = 1

**Solution:**

We have,

(dy/dx) = 1 + x

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

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

^{2})(1 + y^{2})On integrating both sides

tan

^{-1}y = x + (x^{3}/3) + cPut x = 0, y = 1 in above equation

tan

^{-1}(1) = 0 + 0 + cc = π/4

tan

^{-1}y = x + (x^{3}/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 + y^{2 }+ xy^{2}, y(0) = 0

**Solution:**

We have,

(dy/dx) = 1 + x + y

^{2 }+ xy^{2 }(dy/dx) = (1 + x) + y

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

^{2})On integrating both sides

tan

^{-1}(y) = x + (x^{2}/2) + cPut x = 0, y = 0 in above equation

tan

^{-1}(0) = 0 + 0 + cc = 0

tan

^{-1}(y) = x + (x^{2}/2)y = tan(x + x

^{2}/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)

^{2}log(y + 3)^{3}e

^{(y+2) }= x^{2}(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 + x^{2})(dy/dx) + (1 + y^{2}) = 0, y = 1 at x = 0

**Solution:**

We have,

(1 + x

^{2})(dy/dx) + (1 + y^{2}) = 0(1 + x

^{2})(dy/dx) = -(1 + y^{2})On integrating both sides

tan

^{-1}y = -tan^{-1}x + cPut x = 0, y = 1 in above equation

tan

^{-1}(1) = tan^{-1}(0) + cc = π/4

tan

^{-1}y = π/4 – tan^{-1}xy = tan(π/4 – 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∫[(∫xdx]dx + x

^{2 }+ c-cosy + ysiny – ∫sinydy = x

^{2}logx – ∫xdx + x^{2 }+ c-cosy + ysiny + cosy = x

^{2}logx – x^{2}/2 + x^{2 }+ cPut x = 1, y = 0 in above equation

-1 + 0 + 1 = 0 – (1/2) + 1 + c

c = -(1/2)

ysiny = x

^{2}logx + x^{2}/2 – (1/2)2ysiny = 2x

^{2}logx + x^{2 }– 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 + 1Taking 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 – ∫[∫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) = -4xy^{2}, given that y =1 when x = 0.

**Solution:**

We have,

(dy/dx) = -4xy

^{2 }(dy/y

^{2}) + 4xdx = 0On integrating both sides

∫(dy/y

^{2}) + 4∫xdx = 0-(1/y) + 2x

^{2 }= cPut x = 0, y = 1 in above equation

-1 + 0 = c

c = -1

-(1/y) + 2x

^{2 }= -1(1/y) = 2x

^{2 }+ 1y = 1/(2x

^{2 }+ 1)

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

**Solution:**

We have,

(dy/dx) = e

^{x}sinxdy = e

^{x}sinxdxOn integrating both sides

∫dy = ∫e

^{x}sinxdxLet, I = ∫e

^{x}sinxdxI = e

^{x}∫sinx – ∫[∫sinxdx]dxI = -e

^{x}cosx + ∫e^{x}cosxdxI = -e

^{x}cosx + e^{x}∫cosxdx – ∫[∫cosxdx]dxI = -e

^{x}cosx + e^{x}sinx – ∫e^{x}sinxdxI = -e

^{x}cosx + e^{x}sinx – I2I = -e

^{x}cosx + e^{x}sinxI = e

^{x}(sinx – cosx)/2y = 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|x

^{2}(y + 2)^{2}|Curve is passing through (1, -1)

-1 – 1 – c = log(1)

c = 2

y – x – 2 = log|x

^{2}(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)πr

^{3}According to the question (dv/dt) = k

(4/3)π.3r

^{2}(dr/dt) = k4πr

^{2}dr = kdtOn integrating both sides

∫4πr

^{2}dr = ∫kdt4π(r

^{3}/3) = kt + c4πr

^{3 }= 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

4πr

^{3 }= 3(84πt + 36π)r

^{3 }= 63t + 27r = (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 = x

_{0}(x_{0}is numbers of bacteria at t = 0)log(x

_{0}) = 0 + cc = log(x

_{0})On putting the value of c in equation (ii)

log(x) = kt + log(x

_{0})log(x/x

_{0}) = kt -(iii)The number is increased by 10% in 2 hours.

x = x

_{0}(1 + 10/100)(x/x

_{0}) = (11/10)On putting the value of (x/x

_{0}) & t = 2 in equation (iii)2 × k = log(11/10)

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

Therefore, equation (iii) becomes

log(x/x

_{0}) = (1/2)log(11/10) × tAt time t

_{1}numbers of bacteria becomes 200000 from 100000(i.e, x = 2x_{0})t

_{1 }t

_{1}

### Question 58. If y(x) is a solution of the differential equation , and y(0) = 1, then find the value of y(π/2).

**Solution:**

We have,

(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)