Question

Pls help:)
Solve each of the following:
\(a)4y''+4y'+37y=5xe^{-2x}\)
\(b)y"+4y=12x^2-16xcos2x\)

Answer 1

For the a) part I found the complimentary function this way.
Given
\((4D^2+4D+37)y=5xe^{-2x}\)
In auxiliary equation is
\(4\lambda ^2+4\lambda +37=0\)
\[\lambda=\frac{-1}{2}\pm3i\]
Thus, the complimentary function is
\[y_n=e^{\frac{-1x}{2}}(C_1cos3x+C-2sin3x)\]

Answer 2

@wio hav I done it correctly?.

Answer 3

I'm not sure actually.

Answer 4

oh k. Are there any other methods for doing this

Answer 5

It's been a while sense I've done 2nd Order with complex roots.

Answer 6

By complementary solution, you mean the homogeneous solution where we set it equal to 0?

Answer 7

ya.:)

Answer 8

Well, the \(C-2\) threw me off for a while, but realizing it is \(C_2\), I'd say you are correct.

Answer 9

Next is the particular solution.

Answer 10

ya. that's where I am stuck.

Answer 11

Yeah, actually normally I would say guess and check, but I've recently learned there may be a more elegant method.

Answer 12

http://math.mit.edu/suppnotes/suppnotes03/o.pdf

Answer 13

hmm. i did a similar question with real roots. but am finding it hard to do it with the complex one.

Answer 14

Hmmm, how did it go with real roots?

Answer 15

I really don't see a difference, since you just add the homogeneous solution to the particular one.

Answer 16

\(y""-3y'+2y=xe^x\)
this is the question.
The particular solution was found this way.
\[Y_p=\frac{1}{D^2-3D+2}xe^x\]
\[=\frac{1}{(D-1)(D-2)}xe^x\]
\[=\frac{1}{(D-1)}e^{2x}\int e^{-2x}xe^xdx\]
\[=\frac{1}{(D-1)}e^{2x}[-e^{-x}-\int e^{-x}dx\]
\[=\frac{1}{(D-1)}[-(x+1)e^x]\]
\[=-e^x\int e^x(x+1)e^xdx\]
\[=\frac{-1}{2}(1+x)^2e^x\]

Answer 17

So you didn't use undetermined coefficients... interesting.

Yet this method would be tricky for complex roots since it can't be properly factored.

Answer 18

The undetermined coefficients method is to try: \[
Ce^{-2x}(Ax+B)
\]and then solve for the coefficients.

Answer 19

To be honest, I never learned your method. But I think it could be extended with Euler's formula. If I understood it.

Answer 20

Looks like the pattern is: \[
\frac{1}{D-a}f(x) = e^{ax}\int e^{-ax}f(x)dx
\]

Answer 21

ohh. Actually I missed this lecture. somehw i understood this nd can do for questions with real roots. but am unable to do for the complex ones. ya that's the general formula given to us.

Answer 22

I noticed that on the final step with \(D-1\) the inner \(e^x\) doesn't have the negative, but according to the pattern it should...?

Answer 23

@ajprincess Can you try just doing it with complex number numbers?

Answer 24

Just let \(a\) be complex. Let's see the final result I guess.

Answer 25

Really sorry.probs in the internet connection. in the same way as I did for real roots.

Answer 26

@wio do I hav to do it in the same way?

Answer 27

If I were you I would look up the pattern in the book
I would look it up myself if I knew the book and where you were.

Answer 28

And if you find the pattern in the book, post it here so we can learn.

Answer 29

Ha k. sure:)

Answer 30

I found something interesting that may help.

Answer 31

\[
\frac{1}{f(D)}e^{ax}f(x) = e^{ax}\frac{1}{D+a}f(x)
\]

Answer 32

If we shifted it over, it might end up having real roots

Answer 33

http://www.solitaryroad.com/c657.html

Answer 34

do u mean this @wio?