Solving y'=y-sin(2x): Integral of -sin(2x)e^x Stuck, Help!

  • Thread starter Thread starter koolrizi
  • Start date Start date
koolrizi
Messages
21
Reaction score
0
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
 
Physics news on Phys.org
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.
 
directly? I think i am not seeing something
 
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
 
OH ok i got it...wasnt seeing that lol...THanks
 
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)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top