## anonymous one year ago compute y(2) if y(0)=y'(0)=0 and y''-3y'+2y=xe^x

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1. anonymous

@SithsAndGiggles @beginnersmind @IrishBoy123 @Loser66 some of you can help me with this problem? i'll be thankful

2. IrishBoy123

i'm in i'd give it a good hard slap with a Laplace Transform, but there are probably other ways too.

3. anonymous

i tried using the homogenous solution and the particular solution but i can't figure it out

4. IrishBoy123

so first you solved $y''-3y'+2y=0$ and you got an answer?

5. IrishBoy123

then you turned your mind to $y''-3y'+2y=x \ e^x$

6. IrishBoy123

for that particular solution, i'd try: $y = (Px^2 + Qx + R) e^x$ normally: $$y = (Px + Q) e^x$$ would be good but you already have an $$e^x$$ term in the complementary solution

7. anonymous

i don't know if it is right or not

8. IrishBoy123

has this gone a bit skewy at the end? in terms of the IV's you should be putting those into the whole solution, not just the particular, unless i am misreading in terms of method, this is "Variation of Parameters"(??), because there is a really streamlined process on Paul's Online [together with a proof] it looks a lot like what you have done but with fewer steps so the Wronskian $\left|\begin{matrix}e^{2x} & e^x \\ 2e^{2x} & e^x\end{matrix}\right| = - e^{3x}$ which gives $y_p = -e^{2x} \int \frac{e^x.xe^x }{ -e^{3x} } \ dx + e^{x} \int \frac{e^{2x}.xe^x }{ -e^{3x} } \ dx$ which i reckon comes out as $y_p = -xe^x - \frac{x^2e^x}{2}-e^x$ so $y = -xe^x - \frac{x^2e^x}{2}-e^x + A e^{2x} + Be^x$ and put IV's in there for A = 1, B = 0 i've done a transform and i get the same thing link to Paul: http://tutorial.math.lamar.edu/Classes/DE/VariationofParameters.aspx

9. IrishBoy123

another way, a bit quicker

10. anonymous

Your writing is a little messy!

11. IrishBoy123

it's deliberate! it's all a load of rubbish really, i just made it up and posted it :-))

12. anonymous

@IrishBoy123 all you done there is the solve or just a part of it?

13. IrishBoy123

attachment is solution using transform, comments above are just a few comments on the way you are doing it. the solutions on Paul's seem very compact.

14. anonymous

thank you for your time, now it's clear to me. hope i will pass the exam

15. IrishBoy123

if you're bothered, cover the solutions to the examples on Paul's Online and try them yourself good luck with everything!!!