1. Inverse operators

Hey I need to solve this problem by means of inverse operators:
y''-3y'+2y=x
Thank you!

2. Originally Posted by deltafee
Hey I need to solve this problem by means of inverse operators:

y''+3y'+2y=x
Thank you!
Just FYI your questions are very vague.
I think this is what you mean...

$\displaystyle \displaystyle y''+3y'+2y=x \iff (D^2+3D+2)y=x$
Solving for y gives

$\displaystyle \displaystyle y=\frac{1}{(D+1)(D+2)}[x]$

Since $\displaystyle x$ is a polynomial we can express each operator as a geometric series, but 1st partial fractions gives

$\displaystyle \displaystyle \frac{1}{(D+1)(D+2)}=\frac{1}{D+1}+\frac{-1}{D+2}$
Then
$\displaystyle \displaystyle \frac{1}{1+D}=\sum_{n=0}^{\infty}(-1)^nD^n$
and
$\displaystyle \displaystyle \frac{1}{2+D}=\frac{1}{2}\frac{1}{1+\frac{D}{2}}=\ sum_{n=0}^{\infty}(-1)^n\frac{D^n}{2^n}$

Combing the sums gives

$\displaystyle \displaystyle \frac{1}{(D+1)(D+2)}=\sum_{n=0}^{\infty} \left( \frac{2^n+1}{2^n}\right)(-1)^nD^n$

Note that since the right hand side is only a degree 1 polynomial we only need two terms from the series

$\displaystyle \displaystyle 1-\frac{3}{2}D$

$\displaystyle \displaystyle \left( 1-\frac{3}{2}D\right)x=x-\frac{3}{2}$

So this gives $\displaystyle \displaystyle y_p=x-\frac{3}{2}$

You still need to use the eigenfunctions to find the complimentary solution.

3. Very elegant!