D Operator

**khanim** · September 15th 2009, 04:49 AM

Need to find the Particular Integral

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

**HallsofIvy** · September 15th 2009, 05:09 AM

Originally Posted by khanim

Need to find the Particular Integral

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

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

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

Or $D^2y+ 2Dy+ y= 65e^x sin(2x)$ ?

**khanim** · September 15th 2009, 05:15 AM

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

Sorry about that, I have edited the question, I was missing the $y$ after the two

**halbard** · September 15th 2009, 11:54 AM

$y=\frac1{D^2+2D+2}65\mathrm e^x\sin 2x=65\mathrm e^x\frac1{(D+1)^2+2(D+1)+2}\sin 2x$ $=65\mathrm e^x\frac1{D^2+4D+5}\sin 2x$

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

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

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