Solve the ODE

y''''-3y'''+4''=2(x-1)e^x

Question

Solve the ODE

y''''-3y'''+4''=2(x-1)e^x

anonymous · Answer

http://www.wolframalpha.com/input/?i=y%27%27%27%27-3y%27%27%27%2B4y%27%27+%3D+2%28x-1%29e^x

anonymous · Answer

lol ya ok... could've done that... 
need the work

anonymous · Answer

what is 4"

anonymous · Answer

This is a massive question to do online.  I'll do as much as I can...

anonymous · Answer

Make the substitution $$v=y'$$Then your equation becomes$$v''-3v'+4v=e^x(2x-2)$$

anonymous · Answer

You need to solve the homogeneous equation first, then find the particular solution.

anonymous · Answer

$$v''-3v'+4v=0$$has characteristic equation (assuming a solution $$v=e^{{\lambda}x}$$)$${\lambda}^2-3{\lambda}+4=0\rightarrow {\lambda}=\frac{3}{2}{\pm}\frac{\sqrt{7}}{2}i$$which yields a homogeneous solution,$$v=c_1e^{3x/2}\cos(\frac{\sqrt{7}x}{2})+c_2e^{3x/2}\sin(\frac{\sqrt{7}x}{2})$$

anonymous · Answer

The particular solution can be found by attempting a trial solution based on the RHS of the DE, so attempt$$v=e^x(ax+b)$$When you substitute this into the DE, you end up with the following$$2ax+(2b-a)=2x-2$$(the exponentials cancel).  This is true only for a=1 and b=-1/2, so the particular solution is$$v=e^x(x-\frac{1}{2})$$

anonymous · Answer

Your total solution is therefore,

anonymous · Answer

$$v=c_1e^{3x/2}\cos(\frac{\sqrt{7}}{2}x)+c_2e^{3x/2}\sin(\frac{\sqrt{7}}{2}x)+e^x(x-\frac{1}{2})$$

anonymous · Answer

Now, since $$v=\frac{dy}{dx}$$ you have to integrate v to find y...this I shall leave to you...if you need further assistance, let me know.  Good luck.