Answer to Question #88757 in Differential Equations for Mounika

Question

Solve p2 + 2py cotx = y2.

Answer 1

$$(y')^2+2y'ycotx-y^2=0$$

$$((y')^2+2y'ycotx-y^2)/y^2=0$$

$$(y'/y)^2+2y'cotx/y-1=0$$

$$z=y'/y$$

$$z^2+2zcotx-1=0$$

$$D=(2cotx)^2+4=4(1+(cotx)^2)=4/(sinx)^2$$

So we have two cases.

1)

$$z=(-2cotx+\sqrt{4/(sinx)^2})/2=(1-cosx)/sinx$$

$$y'/y=(1-cosx)/sinx$$

$$dy/y=(1-cosx)dx/sinx$$

$$\intop dy/y=\intop (1-cosx)dx/sinx$$

$$lny=\intop (1-cosx)dx/sinx$$

$$\intop (1-cosx)dx/sinx=-\intop (1-cosx)d(cosx)/(sinx)^2=$$

$$=-\intop (1-cosx)d(cosx)/(1-(cosx)^2)=-\intop d(cosx)/(1+cosx)=$$

$$=-ln(1+cosx)+C$$

So we get:

$$lny=-ln(1+cosx)+C=ln(1/(1+cosx))+C$$

$$y(x)=c/(1+cosx)$$

2)

$$z=(-2cotx-\sqrt{4/(sinx)^2})/2=-(1+cosx)/sinx$$

$$lny=-\intop (1+cosx)dx/sinx=-ln(1-cosx)+C$$

$$lny=ln(1/(1-cosx)+C$$

$$y(x)=c/(1-cosx)$$