# Solving the initial value first order differential equation

• Dec 5th 2009, 02:15 PM
genlovesmusic09
Solving the initial value first order differential equation
$\cos (x)y' + \sin (x)y = 2\cos (x)\sin (x) - 1
$

$y(\frac{\pi }
{4}) = 3\sqrt 2 ,0 \leqslant x \leqslant \frac{\pi }
{4}
$

I've tried to get y' and y by itself and got this mess:
$\int {\frac{{dy}}
{y}} = \int {\frac{{2{{\cos }^3}(x)\sin (x) - 1 - \sin (x)dx}}
{{\cos (x)}}}
$
• Dec 5th 2009, 02:19 PM
mr fantastic
Quote:

Originally Posted by genlovesmusic09
$\cos (x)y' + \sin (x)y = 2\cos (x)\sin (x) - 1
$

$y(\frac{\pi }
{4}) = 3\sqrt 2 ,0 \leqslant x \leqslant \frac{\pi }
{4}
$

I've tried to get y' and y by itself and got this mess:
$\int {\frac{{dy}}
{y}} = \int {\frac{{2{{\cos }^3}(x)\sin (x) - 1 - \sin (x)dx}}
{{\cos (x)}}}
$

The DE can be re-written as $\frac{dy}{dx} + (\tan x) y = 2 \sin x - \frac{1}{\cos x}$.

Use the integrating factor technique to solve it.
• Dec 5th 2009, 02:25 PM
genlovesmusic09
where did cos^3(x) go?
I understand how everything else changed
• Dec 5th 2009, 02:39 PM
mr fantastic
Quote:

Originally Posted by genlovesmusic09
where did cos^3(x) go?
I understand how everything else changed

This is the equation you posted (check the OP yourself):
Quote:

$\cos (x)y' + \sin (x)y = 2\cos (x)\sin (x) - 1$
so I have no idea what you mean by
Quote:

where did cos^3(x) go?
• Dec 5th 2009, 02:48 PM
genlovesmusic09
yes I see the confusion. I mistyped the problem it is:
$\cos (x)y' + \sin (x)y = 2{\cos ^3}(x)\sin (x) - 1
$
• Dec 5th 2009, 02:52 PM
mr fantastic
Quote:

Originally Posted by genlovesmusic09
yes I see the confusion. I mistyped the problem it is:
$\cos (x)y' + \sin (x)y = 2{\cos ^3}(x)\sin (x) - 1
$

Well then, all you need to do is make the appropriate small change in my original post and then continue from there.