第 12 类 RD Sharma 解决方案 – 第 22 章微分方程 – 练习 22.5 |设置 1

求解以下微分方程：

问题 1. (dy/dx) = x ² + x – (1/x)

解决方案：

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)

问题 2. (dy/dx) = x ⁵ + x ² – (2/x)

解决方案：

We have,

(dy/dx) = x⁵ + x² – (2/x)     

On integrating both sides, we get

∫(dy) = ∫[x⁵ + x² – (2/x)]dx

y = (x⁶/6) + (x³/3) – 2log(x) + c         -(Here, ‘c’ is integration constant)

问题 3. (dy/dx) + 2x = e ^3x

解决方案：

We have,

(dy/dx) + 2x = e^3x 

(dy/dx) = -2x + e^3x  

dy = [-2x + e^3x]dx

On integrating both sides, we get

∫dy = ∫[-2x + e^3x]dx

y = (-2x²/2) + (e^3x/3) + c

y = (-x²) + (e^3x/3) + c         -(Here, ‘c’ is integration constant)

问题 4. (x ² + 1)(dy/dx) = 1

解决方案：

We have,

(x² + 1)(dy/dx) = 1               

dy = (1/x² + 1)dx

On integrating both sides, we get

∫dy = ∫(dx/x² + 1)

y = tan^-1x + c         -(Here, ‘c’ is integration constant)

问题 5. (dy/dx) = (1 – cosx)/(1 + cosx)

解决方案：

We have,

(dy/dx) = (1 – cosx)/(1 + cosx)     

(dy/dx) = (2sin²x/2)/(2cos²x/2)

dy/dx = tan²(x/2)

dy/dx = [sec²(x/2) – 1]

On integrating both sides, we get

∫dy = ∫[sec²(x/2) – 1]dx

y = [$\frac{1}{\frac{1}{2}}$tan(x/2)] – x + c

y = 2tan(x/2) – x + c         -(Here, ‘c’ is integration constant)

问题 6. (x + 2)(dy/dx) = x ² + 3x + 7

解决方案：

We have,

(x + 2)(dy/dx) = x² + 3x + 7 

(dy/dx) = (x² + 3x + 7)/(x + 2)

(dy/dx) = (x² + 3x + 2 + 5)/(x + 2)

(dy/dx) = [(x + 1)(x + 2) + 5]/(x + 2)

(dy/dx) = (x + 1) + 5/(x + 2)

dy = (x + 1)dx + 5dx/(x + 2) 

On integrating both sides, we get

∫dy = ∫(x + 1)dx + 5∫dx/(x + 2)

y = x²/2 + x + 5log(x + 2) + c         -(Here, ‘c’ is integration constant)

问题 7. (dy/dx) = tan ^-1 x

解决方案：

We have,

(dy/dx) = tan^-1x    

dy = tan^-1xdx

On integrating both sides, we get

∫dy = ∫tan^-1xdx

y = ∫1 × tan^-1xdx

y = tan^-1x∫1dx – ∫[$\frac{d(tan^{-1}x)}{dx}$∫1dx]dx

y = xtan^-1x – ∫\frac{x}{(1 + x2)}dx

y = xtan^-1x – $\frac{1}{2}∫\frac{2x}{1+x^2}dx$

y = xtan^-1x – (1/2)log(1 + x²) + c         -(Here, ‘c’ is integration constant)

问题 8. (dy/dx) = logx

解决方案：

We have,

(dy/dx) = logx  

dy = logxdx

On integrating both sides, we get

∫dy = logx∫1dx – ∫[$\frac{d(logx)}{dx}$∫1dx]dx

y = xlogx – ∫[x/x]dx

y = xlogx – ∫dx

y = xlogx – x + c

y = x(logx – 1) + c         -(Here, ‘c’ is integration constant)

问题 9. (1/x)(dy/dx) = tan ^-1 x

解决方案：

We have,

(1/x)(dy/dx) = tan^-1x 

dy = xtan^-1x dx

On integrating both sides, we get

∫dy = ∫xtan^-1x dx

y = tan^-1x∫xdx – ∫[$\frac{d(tan^{-1}x)}{dx}$∫xdx]dx

y = (x²/2)tan^-1x – $∫(\frac{x^2}{2(1+x^2)})dx$

y = (x²/2)tan^-1x – (1/2)∫1- \frac{1}{(1 + x^2)}dx

y = (x²/2)tan^-1x – (x/2) + (1/2)tan^-1x + c

y = (1/2)(x² + 1)tan^-1x – (x/2) + c         -(Here, ‘c’ is integration constant)

问题 10. (dy/dx) = cos ³ xsin ² x + x√(2x + 1)

解决方案：

We have,

(dy/dx) = cos³xsin²x + x√(2x + 1) 

(dy/dx) = cosxcos²xsin²x + x√(2x + 1)

y = I₁ + I₂

I₁ = cosxcos²xsin²xdx

On integrating both sides, we get

= ∫(1 – sin²x)sin²xcosxdx

Let, sinx = z

On differentiating both sides 

cosxdx = dz

= ∫(1 – z²)z²dz

= z³/3 – z⁵/5 + c₁

= sin³x/3 – sin⁵x/5 + c₁

I₂ = x√(2x + 1)dx

Let, (2x + 1) = u²

On differentiating both sides 

2dx = 2udu

dx = udu

= [(u² – 1)/2]u × udu

I₂ = (1/2)(u⁴-u²)du

On integrating both sides, we get

= (1/2)[u⁵/5 – u³/3] + c₂

= 1/10(2x + 1)^5/2 – 1/6(2x + 1)^3/2 + c₂

y = I₁ + I₂

y = (sin³x/3) – (sin⁵x/5) +  1/10(2x + 1)^5/2 – 1/6(2x + 1)^3/2  + c

问题 11. (sinx + cosx)dy + (cosx – sinx)dx = 0

解决方案：

We have,

(sinx + cosx)dy + (cosx – sinx)dx = 0   

(dy/dx) = -[(cosx – sinx)/(sinx + cosx)]

Let, (sinx + cosx) = z

On differentiating both sides 

(sinx + cosx)dx = dz

dy = (dz/z)

On integrating both sides, we get

∫dy = ∫(dz/z)

y = logz + c

y = log(sinx + cosx) + c

问题 12。 (dy/dx) – xsin ² x = 1/(xlogx)

解决方案：

We have,

(dy/dx) – xsin²x = 1/(xlogx) 

dy = xsin²xdx + 1/(xlogx)dx

On integrating both sides, we get

y = ∫xsin²xdx + ∫dx/(xlogx)

y = I₁ + I₂

I₁ = (1/2)(2sin²x)xdx

= (1/2)[(1 – cos2x)xdx]

= (1/2)(xdx – xcos2x)dx

On integrating both sides, we get

= 1/2[∫xdx – ∫xcos2x dx]

= 1/2(x²/2) – 1/2[x∫cos2x dx – ∫(1∫cos2x dx)]dx

= 1/2(x²/2) – (x/4)sin2x + ∫(1/4sin2x)dx

= 1/2(x²/2) – (x/4)sin2x + ((1/8)cos2x) + c₁

I₂ = 1/(xlogx)dx

Let, logx = z

On differentiating both sides, we get

(dx/x) = dz

= (dz/z)

= logz

= log(logx) + c₂

y = I₁ + I₂

y = (x²/4) – (xsin2x/4) + (cos2x/8) + log(logx) + c

问题 13. (dy/dx) = x ⁵ tan ^-1 (x ³ )

解决方案：

We have,

(dy/dx) = x⁵tan^-1(x³)   

Let, x³ = z

On differentiating both sides, we get

3x²dx = dz

x²dx = dz/3

dy = (1/3)[ztan^-1z]dz

On integrating both sides, we get

∫dy = (1/3)∫ztan^-1z dz

y = (1/3)[tan ^{– 1}z ∫zdz – ∫{$\frac{d(tan ^{-1}z)}{dz}$∫zdz}dz]

y = (z²/6)tan^-1z – $∫(\frac{z^2}{6(1+z^2)})dz$

y = (z²/6)tan^-1z – (1/6)∫1 - \frac{1}{(1 + z^2)}dz

y = (z²/6)tan^-1z – (1/6)∫dz – (1/6)∫dz/(1 + z²)

y = (z²/6)tan^-1z – (z/6) – (1/6)tan^-1z

y = (1/6)(z² + 1)tan^-1z – (z/6) + c     

y = (1/6)[(x⁶ + 1)tan – 1(x³) – (x³)] + c