I'm kind of stuck on this problem. The problem given is cos^{2}(t)y' + 4y = 5 , where y(0) = 6

So what I got are no where near the choices give.

$\displaystyle y' + \frac{4y}{cos^2(t)} = 5cos^2(t)$

$\displaystyle \frac{dy}{dt} + \frac{4y}{cos^2(t)} = 5cos^2(t)$

$\displaystyle e^{\int\frac{4}{cos^2(t)}dt} = e^{4tan(t)}$

$\displaystyle e^{4tan(t)} \frac{dy}{dt} + e^{4tan(t)}\frac{4}{cos^2(t)} y = 5cos^2(t)$

$\displaystyle \int \frac{d}{dt} e^{4tan(t)}y = \int 5cos^2(t)$

$\displaystyle e^{4tan(t)} y = \frac{\frac{5}{2}(t+sin(t)cos(t))+c}{e^{4tan(t)}} $

When I Plug in y(0)=6, I get C = 6....

The choices for the answers are

y(t) = 4+2e^{tan(t)}

$\displaystyle y(t) = \frac{5}{4}+\frac{19}{4}e^{-4tan(t)}$

$\displaystyle y(t) = \frac{5}{4}+\frac{19}{4}e^{-tan(t)}$

y(t) = 4+2e^{-tan(t)}