Question

Solve the differential equation p2+2py cot x-y2=0 , where p=y' .

Answer 1

Answer on question 40471 – Math – Differential equation

Solve the differential equation

$$p^2 + 2 p y \cot x - y^2 = 0,$$

where $p = y'$.

Solution:

We have

$$(y')^2 + 2 y' y \cot x - y^2 = 0,$$ $$\frac{(y')^2 + 2 y' y \cot x - y^2}{y^2} = 0, \qquad y \neq 0,$$ $$\left(\frac{y'}{y}\right)^2 + 2 \frac{y'}{y} \cot x - 1 = 0.$$

Denote

$$z = \frac{y'}{y}.$$

Thus we have next equation

$$z^2 + 2 z \cot x - 1 = 0,$$ $$D = (2 \cot x)^2 - 4 \cdot 1 \cdot (-1) = 4 \cot^2 x + 4 = 4 (1 + \cot^2 x) = \frac{4}{\sin^2 x} > 0.$$

So we get two cases

1)

$$z = \frac{-2 \cot x + \sqrt{\frac{4}{\sin^2 x}}}{2} = -\frac{\cos x}{\sin x} + \frac{1}{\sin x} = \frac{1 - \cos x}{\sin x},$$ $$\frac{y'}{y} = \frac{1 - \cos x}{\sin x},$$ $$\frac{dy}{y} = \frac{1 - \cos x}{\sin x} \, dx,$$ $$\int \frac{dy}{y} = \int \frac{1 - \cos x}{\sin x} \, dx,$$ $$\ln y = \int \frac{1 - \cos x}{\sin x} \, dx.$$

We have

$$\begin{aligned} & \int \frac{1 - \cos x}{\sin x} dx = - \int \frac{1 - \cos x}{\sin^2 x} d(\cos x) = - \int \frac{1 - \cos x}{1 - \cos^2 x} d(\cos x) = \\ & = - \int \frac{1 - \cos x}{(1 - \cos x)(1 + \cos x)} d(\cos x) = - \int \frac{1}{1 + \cos x} d(\cos x) = \\ & = - \int \frac{1}{1 + \cos x} d(1 + \cos x) = - \ln(1 + \cos x) + C. \end{aligned}$$

So we get

$$\ln y = - \ln(1 + \cos x) + C,$$ $$\ln y = \ln \frac{1}{1 + \cos x} + C,$$ $$y(x) = \frac{c}{1 + \cos x},$$

where $c = e^c$.

2)

$$z = \frac{ -2 \cot x - \sqrt{\frac{4}{\sin^2 x} } }{2} = - \frac{\cos x}{\sin x} - \frac{1}{\sin x} = - \frac{1 + \cos x}{\sin x},$$ $$\frac{y'}{y} = - \frac{1 + \cos x}{\sin x},$$ $$\frac{dy}{y} = - \frac{1 + \cos x}{\sin x} dx,$$ $$\int \frac{dy}{y} = - \int \frac{1 + \cos x}{\sin x} dx,$$ $$\ln y = - \int \frac{1 + \cos x}{\sin x} dx.$$

We have

$$\begin{aligned} & - \int \frac{1 + \cos x}{\sin x} dx = \int \frac{1 + \cos x}{\sin^2 x} d(\cos x) = \int \frac{1 + \cos x}{1 - \cos^2 x} d(\cos x) = \\ & = \int \frac{1 + \cos x}{(1 - \cos x)(1 + \cos x)} d(\cos x) = - \int \frac{1}{1 - \cos x} d(1 - \cos x) = \\ & = - \int \frac{1}{1 - \cos x} d(1 - \cos x) = - \ln(1 - \cos x) + C. \end{aligned}$$

So we get

$$\ln y = - \ln (1 - \cos x) + C,$$ $$\ln y = \ln \frac {1}{1 - \cos x} + C,$$ $$y (x) = \frac {c}{1 - \cos x},$$

where $c = e^{C}$

Answer:

$$y (x) = \frac {c}{1 + \cos x}$$

or

$$y (x) = \frac {c}{1 - \cos x}.$$