Question

find solution of differential equation. I really need help, not sure what to do!!

4y''-y=xe^(x/2)

Answer 1

I have the general solution, i just dont know how to get the particular solution with variation of parameters

Answer 2

Hi! I'm just learning ODE now! But I know you can't get a particular solution without given conditions.

You'd need something like
\(y(t_0)=y_0\qquad\)and\(\qquad y'(t_0)=y_0'\)

Answer 3

So the general solution would be what you want, I think. It would be a "family of solutions," meaning a bunch of solutions that are related. Get it? Related... Family.. Hehe... Anyway, you would end up with constants. And you'd have to have values for them for the solution to be specific.

Answer 4

Does that sound about right?

Answer 5

I got the homogeneous solution: \[y=c_{1}e ^{x/2}+c _{2}e ^{-x/2}\]
I just need the particular solution to satisfy (xe^(x/2))/4. I do not need initial values for this problem

Answer 6

Ohh... Okay, I haven't learned this yet. This looks like characteristic equation or principle of superposition stuff, but I don't know if I can handle this problem, sorry!

Answer 7

Why don't you take the derivatives of \(y\) and plug them in?

Answer 8

do you know of anyone who could help?

Answer 9

That didn't help.. Here's my work... It didn't help find a specific solution.
\(y=c_{1}e ^{x/2}+c _{2}e ^{-x/2}\)

\(y'=\frac{1}{2}c_1e^{x/2}-\frac 1 2 c_2e^{-x/2}\)

\(y''=\frac{1}{4}c_1e^{x/2}+\frac 1 4 c_2e^{-x/2}\)

Since \(4y''-y=xe^{x/2}\), we substitute in \(y\) and \(y''\) to find that
\(4\left(\frac{1}{4}c_1e^{x/2}+\frac 1 4 c_2e^{-x/2}\right)-\left(c_{1}e ^{x/2}+c _{2}e ^{-x/2}\right)=xe^{x/2}\)

\(\left(c_1e^{x/2}+c_2e^{-x/2}\right)-\left(c_{1}e ^{x/2}+c _{2}e ^{-x/2}\right)=xe^{x/2}\)

\(c_1e^{x/2}+c_2e^{-x/2}-c_{1}e ^{x/2}-c _{2}e ^{-x/2}=xe^{x/2}\)

\(0=xe^{x/2}\)

Answer 10

do you know of anyone who could help??

Answer 11

I messaged somebody on here!

Answer 12

can you direct them to this problem. I need to turn it in soon. running out of time, AH!

Answer 13

I gotcha!

Answer 14

What process do you solve it by? If \(xe^{x/2}\) was \(0\), I would use the characteristic equation stuff.

Answer 15

The problem requires Variation of Parameters.

Answer 16

But I don't think the characteristic equation stuff applies if that's not a 0.

Answer 17

http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx

Answer 18

is y a function of x or t?

Answer 19

it does, there are just additional steps (wronskian...) to solve for the particular solution that satisfies the nonhomogeneous part on the right hand side.

Answer 20

Haha, I put \(t\)'s there out of habit... Must be \(x\)!

Answer 21

x

Answer 22

k so the other side

Answer 23

Okay, I will be getting to the Wronskian stuff Friday or next week... Not in time for this... Unless I learn very quickly.

Answer 24

Wronski isn't bad, it's just a matrix to test LI

Answer 25

so sshot, try and test your solution, is it LI? I'm guessing not, because you need an \(x*e^{x/2}\), which you can get from them not being LI and adding in the x. Then you should get a good answer

Answer 26

Check example 5 from that link

Answer 27

Ok, I think i got it now.

Answer 28

np, best of luck!

Answer 29

Thanks, @FibonacciChick666 !