Question

Find a particular solution of y"-2y'+y=-e^(x)[(3+4x-x^2)cosx+(3-4x-x^2)sinx].

Answer 1

@SolomonZelman

Answer 2

Roots of auxiliary equation: \(\color{#000000}{ \displaystyle r=1 }\) (Repeated, w/ multiplicity s=2).

(a policy-related request:
Please, in the future show most work that you can. That means
show the homogeneous solution or at least the roots of the
characteristic/auxiliary equation. Don't just give a blank query.)

So, then the fundimental solution set for homoegeneous solution is:
\(\color{#000000}{ \displaystyle S=\left\{e^x,xe^x\right\} }\)
(if you ever get a linearly dependent S, then you are wrong!!)

and the homogeneous solution then is:
\(\color{#000000}{ \displaystyle y_h(x)=c_1e^x+c_2xe^x }\)

Answer 3

Your initial guess is:
\(\color{#000000}{ \displaystyle y_p=e^x\left(Ax^2+Bx+C\right) \left(D\sin x+\cos x\right) }\)

Answer 4

My internet crushed on me when I had 2 pages of text with codes, so all I am going to say that when you do this correctly you should end up with

\(\color{#000000}{ A=-1 }\)
\(\color{#000000}{ B=0 }\)
\(\color{#000000}{ C=1 }\)
\(\color{#000000}{ D=1 }\)

Answer 5

Also, note for the initial guess.

Your initial guess is not \(\color{#000000}{ \displaystyle Ae^x\left(Bx^2+Cx+D\right) \left(E\sin x+F\cos x\right) }\)
for the following reason.

Does it matter if sine-terms are F/E times greater than cosine-terms, or if sine-terms are just E times greater than cosine terms? You would agree that it does not, right?

You already have all your generic coeffients in your generic quadratic term, so would it suffice if you have 3 unknown coefficients for the quadratic, or do you need to multiply each one of them by A? You would agree that multiplying \(e^x\) at the beginning by A is a waste (although certainly not "wrong").

Answer 6

By quickly noticing this, my guess simplified to
\(\color{#000000}{ \displaystyle y_p(x)=e^x(Ax^2+Bx+C)(D\sin x+\cos x) }\)

Answer 7

G••d Luck!

Answer 8

LOL, I finally got the right answer! It's because I made a mistake in the middle and I did the problem again. Thanks a lot for the help!