Question

Variable Seperable

Cos² y dx + Sin² x dy = 0; y (π/4) =(π/4)

Answer 1

Let us solve the differential equation $\cos^2y\ dx + \sin^2 x\ dy = 0$ which is equivalen to $\sin^2 x\ dy = -\cos^2y\ dx.$

If $\sin x=0,$ then $x=\pi n,\ n\in\Z,$ are the solutions of the differential equation $\cos^2y\ dx + \sin^2 x\ dy = 0$. Also if $\cos y=0,$ then $y=\frac{\pi}{2}+\pi n,\ n\in\Z,$ are the solutions of the differential equation $\cos^2y\ dx + \sin^2 x\ dy = 0$. Since $y(\frac{\pi}{4})\ne\frac{\pi}{4},$ we conclude that they are not the solutions of $\cos^2y\ dx + \sin^2 x\ dy = 0, \ y(\frac{\pi}{4})=\frac{\pi}{4}.$

Let us divide both parts of this equation by $\sin^2x\cos^2y.$ Then we get the following differential equation:

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

Then

$\int\frac{dy}{\cos^2 y}=-\int\frac{dx}{\sin^2 x}$ and hence

$\tan y=\ctg x+C.$

Since $y(\frac{\pi}{4})=\frac{\pi}{4},$ we have that $\tan \frac{\pi}{4}=\ctg \frac{\pi}{4}+C.$ It follows that $1=1+C,$ and hence $C=0.$

We conclude that $\tan y=\ctg x=\tan(\frac{\pi}{2}-x).$ It follows that $y=\frac{\pi}{2}-x+\pi n,\ n\in\Z,$ is the family of solutions of the last differential equation. Taking into account that $y(\frac{\pi}{4})=\frac{\pi}{4},$ we conclude that $y=\frac{\pi}{2}-x$ is the solution of the differential equation $\cos^2y\ dx + \sin^2 x\ dy = 0, \ y(\frac{\pi}{4})=\frac{\pi}{4}.$