# Solving y'=y-sin(2x)

1. Jan 22, 2007

### koolrizi

I found the integrating factor to be e^x for DE y'=y-sin(2x)
Now i am stuck at integral of -sin(2x)e^x
Can you help me with it? I tried using integration by parts and i get to following
integral(sin(2x)(e^x)=sin(2x)e^x-2cos(2x)e^x-4*integral(sin(2x)(e^x)

I am stuck. Help! Thanks

2. Jan 22, 2007

### Manchot

Notice that you have the expression "integral(sin(2x)(e^x))" on both sides of your equation. You can therefore solve for it directly, and be done.

3. Jan 23, 2007

### koolrizi

directly? I think i am not seeing something

4. Jan 23, 2007

### koolrizi

The answer to this DE is 1/5(sin(2x)+2cos(2x) and i am not seeing how to get to this from what i have

5. Jan 23, 2007

### koolrizi

OH ok i got it....wasnt seeing that lol...THanks

6. Jan 23, 2007

### joob

also careful with your results. looks like you missed a couple small details. your integrating factor is actually $e^{-x}$ and you should have the boundaries of the LHS of your equation(unless they're supposed to be =0)