第12类RD Sharma解–第22章微分方程–练习22.6

问题1.解决以下微分方程

$\frac{dv}{dx}+\frac{1+y^2}{y}=0$

解决方案：

From the question it is given that,

$\frac{dy}{dx} + \frac{(1 + y^2)}{y} = 0$

Transposing we get,

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

By cross multiplication,

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

Integrating on both side, we will get,

$∫\frac{y}{(1 + y^2)} dy = ∫-dx$

$∫\frac{2y}{(1 + y^2)} dy = -2 ∫dx$

log (1 + y²) = – 2x + c₁

Therefore, $\frac{1}{2}$log [1 + y²] + x = c

问题2.解决以下微分方程

$\frac{dy}{dx} = \frac{(1 + y^2)}{y^3}$

解决方案：

From the question it is given that,

$\frac{dy}{dx} = \frac{(1 + y^2)}{y^3}$

By cross multiplication,

$\frac{y^3}{(1 + y^2)} dy = dx$

Integrating on both side, we will get,

$∫(y - \frac{y}{(1 + y^2)} dy = ∫dx$

$∫ydy - ∫\frac{y}{(1 + y^2)} dy = ∫ dx\\ ∫ydy - \frac{1}{2} ∫\frac{2y}{(1 + y2)} dy = ∫ dx\\ \frac{y^2}{2} - \frac{1}{2} log [y^2 + 1] = x + c$

问题3.解决以下微分方程：

$\frac{dy}{dx} = sin^2 y$

解决方案：

From the question it is given that,

$\frac{dy}{dx} = sin^2 y$

By cross multiplication,

$\frac{dy}{sin^2 y} = dx$

As we know that, $\frac{1}{sin x}$ = cosec x

cosec²y dy = dx

Integrating on both side, we will get,

∫cosec² y dy = ∫dx + c

– cot y = x + c

问题4.解决以下微分方程：

$\frac{dy}{dx} = \frac{(1 - cos 2y)}{(1 + cos 2y)}$

解决方案：

From the question it is given that,

$\frac{dy}{dx} = \frac{(1 - cos 2y)}{(1 + cos 2y)}$

We know that, 1 – cos 2y = 2sin²y and 1 + cos 2y = 2 cos²y

So, $\frac{dy}{dx} = \frac{(2 sin^2 y)}{(2 cos^2 y)}$

Also we know that, $\frac{sin θ}{cos θ}$ = tan θ

By cross multiplication,

$\frac{dy}{tan^2 y} = dx$

Integrating on both side, we get,

∫cot²y dy = ∫dx

∫ (cosec²y – 1) dy = ∫dx

– cot y- y + c = x

c = x + y + cot y