Question

solve

solve dy/dx = 2( cos^2x) - sinx^2 y^2 /2cosx ;y(0)=-1

Answer 1

Here problem is not typed properly.

I think problem will be

$\frac{dy}{dx}= \frac{(2cos²x-sin²x)y²}{2cosx}$

$\frac{dy}{y²}= \frac{(2cos²x-sin²x)}{2cosx}dx$

Integrating both sides

$\int\frac{dy}{y²}= \int\frac{(2cos²x-sin²x)}{2cosx}dx$

$=> \frac{dy}{y²}= \frac{(3cos²x-1)}{2cosx}dx$

=> $\int\frac{dy}{y²}= \int\frac{3}{2}cosxdx-\int\frac{1}{2} secxdx$

=> $\frac{-1}{y}= \frac{3}{2}sinx-\frac{1}{2} ln|secx+tanx| + C$

By given initial condition, when x = 0, y = -1

1 = 0 - 0 + C

So C = 1

So the particular solution is

$\frac{-1}{y}= \frac{3}{2}sinx-\frac{1}{2} ln|secx+tanx| + 1$

=> 3ysinx - y ln|secx + tanx| + 2y + 2 = 0