Question

compute y(2) if y(0)=y'(0)=0 and y''-3y'+2y=xe^x

Answer 1

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

Answer 2

i'm in

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

Answer 3

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

Answer 4

so first you solved
\[ y''-3y'+2y=0\]

and you got an answer?

Answer 5

then you turned your mind to

\[y''-3y'+2y=x \ e^x\]

Answer 6

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

Answer 7

i don't know if it is right or not

Answer 8

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

Answer 9

another way, a bit quicker

Answer 10

Your writing is a little messy!

Answer 11

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

Answer 12

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

Answer 13

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.

Answer 14

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

Answer 15

if you're bothered, cover the solutions to the examples on Paul's Online and try them yourself

good luck with everything!!!