Question

Solve the following differential equation:

`(cos^2y)/x dy + (cos^2x)/y dx` = 0

Answer 1

`(cos^2y)/x dy + (cos^2x)/y dx` = 0

∴ y cos²y dy + x cos² x dx = 0

∴ `x((1 + cos2x)/2) dx + y((1 + cos 2y)/2) 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

`int xdx + int y dy + int x cos 2x dx + int y cos 2y dy = c_1`      .....(i)

Using integration by parts

`int x cos 2x dx = x int cos 2x dx - int [d/dx (x) int cos 2x dx]dx`

= `x((sin 2x)/2) - int 1. (sin 2x)/2 dx`

= `(x sin 2x)/2 + 1/2 . (cos 2x)/2`

= `(x sin 2x)/2 + (cos 2x)/4`

Similarly, `int y cos 2y dy = (y sin 2y)/2 + (cos 2y)/4`

∴ From equation (i), we get

`x^2/2 + y^2/2 + (x sin 2x)/2 + (cos 2x)/4 + (y sin 2y)/2 + (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.