I am just learning about inverse operators in solving a differential equation, but I don't understand exactly how they work. For example, find a particular solution to
$$4y''-3y'+9y=5x^2$$
using inverse operators.
The above equation is equivalent to
$$(4D^2-3D+9)y=5x^2$$
Now the way to solve this would be to use the inverse operator as follows:
$$y_p=(4D^2-3D+9)^{-1}5x^2$$
or
$$y_p=\frac {1}{4D^2-3D+9}5x^2$$
The book I am reading uses "simple division" and arrives at a result, but I don't quite understand how that works.
What is a step-by-step method to solve the above problem?
Best Answer
Notice that $$4D^2 - 3D + 9 = \left [D - \left ( \frac{3}{8} (1 - i \sqrt{15} )\right)\right ] \left [D - \left ( \frac{3}{8} (1 + i \sqrt{15} )\right)\right ] $$ Also note that $$ \frac{1}{-a( 1 - D /a )} = -\frac{1}{a} \left ( 1 + \frac{D}{a} + \frac{D^2}{a^2} + \ldots \right) $$ Thus to find $y_p$, we simply have to compute ( calling the roots $\lambda_\pm = 3/8 ( 1 \pm i \sqrt{15})$) $$ y_p = \frac{1}{4D^2 - 3D + 9} 5x^2 = \frac{1}{ | \lambda_\pm|^2} \frac{1}{1- D/\lambda_+} \frac{ 1}{1 - D / \lambda_-} 5x^2= \frac{1}{|\lambda_\pm|^2}\left ( 1 + \frac{D}{\lambda_+} + \frac{D^2}{\lambda_+^2} \right ) \left ( 1 + \frac{D}{\lambda_-} + \frac{D^2}{\lambda_-^2} \right) 5x^2 = \frac{5}{|\lambda_\pm|^2} \left ( 1+ \frac{ \lambda_+ + \lambda_-}{|\lambda_\pm|^2} D + \frac{(\lambda_- + \lambda_+)^2} {|\lambda_\pm|^4 } D^2 \right ) x^2$$ Now you just have to evaluate the derivatives and you're done. (where $|\lambda_\pm|^2 = \lambda_+ \lambda_-$)
Note that I stopped at the second differential operator since $D^3 ( 5x^2) = 0$.