# Math Help - cos^2(t)y'+4y=5

1. ## cos^2(t)y'+4y=5

I'm kind of stuck on this problem. The problem given is cos2(t)y' + 4y = 5 , where y(0) = 6

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

$y' + \frac{4y}{cos^2(t)} = 5cos^2(t)$
$\frac{dy}{dt} + \frac{4y}{cos^2(t)} = 5cos^2(t)$
$e^{\int\frac{4}{cos^2(t)}dt} = e^{4tan(t)}$
$e^{4tan(t)} \frac{dy}{dt} + e^{4tan(t)}\frac{4}{cos^2(t)} y = 5cos^2(t)$
$\int \frac{d}{dt} e^{4tan(t)}y = \int 5cos^2(t)$
$e^{4tan(t)} y = \frac{\frac{5}{2}(t+sin(t)cos(t))+c}{e^{4tan(t)}}$

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

The choices for the answers are
y(t) = 4+2etan(t)
$y(t) = \frac{5}{4}+\frac{19}{4}e^{-4tan(t)}$
$y(t) = \frac{5}{4}+\frac{19}{4}e^{-tan(t)}$
y(t) = 4+2e-tan(t)

which is it?

4t or 4y?

3. ## Re: cos^2(t)y'+4y=5

ok nm it's clearly 4y

$\cos^2(t)y'+4y=5$

I'm seeing the answer is $y(t)=\frac{1}{4} e^{-4 \tan (t)} \left(5 e^{4 \tan (t)}+19\right)$

but I have no idea yet how that is obtained. Maybe using an integrating factor. One of the serious math guys here can probably shed some light on it.

4. ## Re: cos^2(t)y'+4y=5

Yes, it's an integrating factor type, and that attempt in #1 is along the right lines except that it should be 5sec^2 (t) on the rhs.

5. ## Re: cos^2(t)y'+4y=5

Originally Posted by JC05
I'm kind of stuck on this problem. The problem given is cos2(t)y' + 4y = 5 , where y(0) = 6

So what I got are no where near the choices give.
$\cos^2(t) y'+4y=5$

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

use integrating factor

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

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

$\frac{d}{dt}\left(\mu(t)y\right)=\frac{5 e^{4 \tan(t)}}{\cos^2(t)}$

$\mu(t) y=\int\frac{5 e^{4 \tan(t)}}{\cos^2(t)}dt=\frac{5}{4}e^{4\tan(t)}+C$

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

$y(0)=\frac{5}{4}+C=6 \Rightarrow C=\frac{19}{4}$

$y=\frac{1}{4}\left(5+19e^{-\tan(t)}\right)$

so your third choice is correct.