## anonymous 5 years ago Solve the ODE y''''-3y'''+4''=2(x-1)e^x

1. anonymous
2. anonymous

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

3. anonymous

what is 4"

4. anonymous

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

5. anonymous

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

6. anonymous

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

7. anonymous

$v''-3v'+4v=0$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,$v=c_1e^{3x/2}\cos(\frac{\sqrt{7}x}{2})+c_2e^{3x/2}\sin(\frac{\sqrt{7}x}{2})$

8. anonymous

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$v=e^x(x-\frac{1}{2})$

9. anonymous

10. anonymous

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

11. anonymous

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.