Question

Solve the following differential equation:

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

Answer 1

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

∴ y cos²y dy + x cos² x dx = 0

∴ ${\displaystyle x\left(\frac{1+\cos 2x}{2}\right)dx+y\left(\frac{1+\cos 2y}{2}\right)dy}$ = 0

∴ x(1 + cos 2x) dx + y(1 + cos 2y)dy = 0

∴ x dx + x cos 2x dx + y dy + y cos 2y dy = 0

Integrating both sides, we get

${\displaystyle \int xdx+\int ydy+\int x\cos 2xdx+\int y\cos 2ydy=c_1}$      .....(i)

Using integration by parts

${\displaystyle \int x\cos 2xdx=x\int \cos 2xdx-\int [\frac{d}{dx}\left(x\right)\int \cos 2xdx]dx}$

= ${\displaystyle x\left(\frac{\sin 2x}{2}\right)-\int 1.\frac{\sin 2x}{2}dx}$

= ${\displaystyle \frac{x\sin 2x}{2}+\frac{1}{2}.\frac{\cos 2x}{2}}$

= ${\displaystyle \frac{x\sin 2x}{2}+\frac{\cos 2x}{4}}$

Similarly, ${\displaystyle \int y\cos 2ydy=\frac{y\sin 2y}{2}+\frac{\cos 2y}{4}}$

∴ From equation (i), we get

${\displaystyle \frac{x^2}{2}+\frac{y^2}{2}+\frac{x\sin 2x}{2}+\frac{\cos 2x}{4}+\frac{y\sin 2y}{2}+\frac{\cos 2y}{4}}$ = c₁

Multiplying throughout by 4, this becomes

2x² + 2y² + 2x sin 2x + cos 2x + 2y sin 2y + cos 2y = 4c₁ 

∴ 2(x² + y²) + 2(x sin 2x + y sin 2y) + cos 2y + cos 2x + c = 0

where c = – 4c₁

This is the general solution.