第12类RD Sharma解–第22章微分方程–练习22.7 |套装3

解微分方程(问题40-48)：

问题40. 2x(dy / dx)= 3y，y(1)= 2

解决方案：

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 

问题41. xy(dy / dx)= y + 2，y(2)= 0

解决方案：

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)

问题42.(dy / dx)= 2e ^x y ³ ，y(0)= 1/2

解决方案：

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

问题43.(dr / dt)= -rt，r(0)= r ₀

解决方案：

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 = r₀$e^{-\frac{t^2}{2}}$

问题44.(dy / dx)= ysin2x，y(0)= 1

解决方案：

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 = $e^{sin^2x}$

问题45(i)。 (dy / dx)= ytanx，y(0)= 1

解决方案：

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

问题45(ii)。 2x(dy / dx)= 5y，y(1)= 1

解决方案：

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)

问题45(iii)。 (dy / dx)= 2e ^2x y ² ，y(0)= -1

解决方案：

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

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

解决方案：

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)

问题45(v)。 (dy / dx)= 2xy，y(0)= 1

解决方案：

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

问题45(vi)。 (dy / dx)= 1 + x ² + y ² + x ² y ² ，y(0)= 1

解决方案：

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

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

解决方案：

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

问题45(viii)。 (dy / dx)= 1 + x + y ² + xy ² ，y(0)= 0

解决方案：

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)

问题45(ix)。 2(y + 3)– xy(dy / dx)= 0，y(1)= -2

解决方案：

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

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

解决方案：

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

问题47.(1 + x ² )(dy / dx)+(1 + y ² )= 0，x = 0时y = 1

解决方案：

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

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

解决方案：

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

问题49.在x = 0时y(0)= 3的情况下，求微分方程e ^{(dy / dx)= x + 1的特定解。}

解决方案：

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

问题50.假设x =π/ 2时y =π/ 2，则求微分方程cosydy + cosxsinydx = 0的解。

解决方案：

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

问题51.假设x = 0时y = 1，则求微分方程(dy / dx)= -4xy ^{2的特定解。}

解决方案：

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)

问题52.找到一条通过点(0，0)且其微分方程为(dy / dx)= e ^x sinx的曲线的方程。

解决方案：

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

问题53.对于微分方程xy(dy / dx)=(x + 2)(y + 2)，找到通过点(1，-1)的求解曲线。

解决方案：

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|

问题54：正在膨胀的球形气球的体积以恒定的速率变化。如果最初的半径为3个单位，则3秒钟后为6个单位。在t秒后找到气球的半径。

解决方案：

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 ballon after t second is (63t + 27)^1/3

问题55.在银行中，本金以每年r％的速度增长。如果Rs 100在10年内翻倍，则求r的值(对数2 = 0.6931)。

解决方案：

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

问题56.在银行中，本金每年以5％的速度增长。该银行存入1000卢比，十年后将值多少钱(e = 1.648)。

解决方案：

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

问题57.在一种培养物中，细菌的数量为100000。在2小时内，细菌的数量增加了10％。如果细菌的生长速度与存在的细菌数量成正比，那么计数将在几小时内达到200000?

解决方案：

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$=\frac{2log2}{log\frac{11}{10}}$

问题58.如果y(x)是微分方程的解$[\frac{(2 + sinx)}{(1+y)}](\frac{dy}{dx})=-cosx$ ，并且y(0)= 1，然后找到y(π/ 2)的值。

解决方案：

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)