Integrate 1st Order Diff EQ: Sin2x - 2Cos2x + c

[Quadruple Bypass]

howdy, i uhh solved this problem early in the semester, and now I am looking at it like wtf did i do to get the right answer...it says use the 'off the shelf formula' for first order diff eqns.

its pretty much just integrating now

y(x) = e^(x) (int[e^(-x)*-sin(2x)dx] + c)

ans => y(x)= 1/5(sin2x-2cos2x)+c

thanks in advance

 

[berkeman]

Are you sure you transcribed those correctly? What do you get when you differentiate the answer you've listed? Just sin and cos terms, right?

 

[courtrigrad]

If y(x) = e^{x}\int e^{-x}\times -\sin 2x \; dx

 

[Quadruple Bypass]

ya that's what i don't understand..

 

[courtrigrad]

use integration by parts.

 

[HallsofIvy]

Ignore the e^x outside the integral until you have done the integral itself, using integration by parts (twice) as courtigrad said. (Presumably the integration will have a factor of e^-x, canceling the e^x outside the integral.)