Question

Solve:
$\frac{{\cos }^{2}y}{x}dy+\frac{{\cos }^{2}xdx}{y}=0$

Answer 1

Given,

$\frac{{\cos }^{2}y}{x}dy+\frac{{\cos }^{2}x}{y}dx=0$

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

$y{\cos }^{2}ydy=-x{\cos }^{2}xdx$

integrating on both sides, we get,

$\int y{\cos }^{2}ydy=-\int x{\cos }^{2}xdx$

Applying integration by parts, $u=y,v^{\prime }={\cos }^2\left(y\right),u=x,v^{\prime }={\cos }^2\left(x\right)$

we get,

$\frac{1}{2}y(y+\frac{1}{2}\sin \left(2y\right))-\int \frac{1}{2}(y+\frac{1}{2}\sin \left(2y\right))dy=-(\frac{1}{2}x(x+\frac{1}{2}\sin \left(2x\right))-\int \frac{1}{2}(x+\frac{1}{2}\sin \left(2x\right))dx)$

$\frac{1}{2}y(y+\frac{1}{2}\sin \left(2y\right))-\frac{1}{2}(\frac{y^2}{2}-\frac{1}{4}\cos \left(2y\right))=-(\frac{1}{2}x(x+\frac{1}{2}\sin \left(2x\right))-\frac{1}{2}(\frac{x^2}{2}-\frac{1}{4}\cos \left(2x\right)))$