# D Operator

• Sep 15th 2009, 04:49 AM
khanim
D Operator
Need to find the Particular Integral

$\displaystyle D^{2}y+2Dy+2y=65e^{x}sin 2x$

(Punch)(Punch)
• Sep 15th 2009, 05:09 AM
HallsofIvy
Quote:

Originally Posted by khanim
Need to find the Particular Integral

$\displaystyle D^{2}y+2Dy+2=65e^{x}sin 2x$

(Punch)(Punch)

What you have written would be more simply written as $\displaystyle D^2y+ 2Dy= 65e^x sin(2x)- 2$. Was that what you meant?

Or did you mean $\displaystyle D^2y+ 2Dy+ 2y= 65e^x sin(2x)$?

Or $\displaystyle D^2y+ 2Dy+ y= 65e^x sin(2x)$?
• Sep 15th 2009, 05:15 AM
khanim
$\displaystyle D^{2}y+2Dy+2y=65e^{x}sin 2x$

Sorry about that, I have edited the question, I was missing the $\displaystyle y$ after the two (Sleepy)
• Sep 15th 2009, 11:54 AM
halbard
$\displaystyle y=\frac1{D^2+2D+2}65\mathrm e^x\sin 2x=65\mathrm e^x\frac1{(D+1)^2+2(D+1)+2}\sin 2x$ $\displaystyle =65\mathrm e^x\frac1{D^2+4D+5}\sin 2x$

$\displaystyle {}=65\mathrm e^x\frac1{-2^2+4D+5}\sin 2x=65\mathrm e^x\frac1{4D+1}\sin 2x=65\mathrm e^x\frac{4D-1}{16D^2-1}\sin 2x$

$\displaystyle {}=65\mathrm e^x\frac{4D-1}{16(-2^2)-1}\sin 2x=\mathrm e^x(1-4D)\sin 2x=\mathrm e^x(\sin 2x-8\cos 2x)$.