2nd order de with product in right hand of equation

[Jake89]

hello,

the problem

y''-3y'+2y = e^3x*sinx

confuses me

I have Yh but i need to know how to tackle e^3x*sinx

How do i construct Yp for this functions, and how do i differentiate them?
Would be very thankful if someone could help me to finish this problem and get the constants

 

[galactus]

Use\(\displaystyle y_{p}=Ae^{3x}cos(x)+Be^{3x}sin(x)\)

\(\displaystyle y'=(3A+B)e^{3x}cos(x)+(3B-A)e^{3x}sin(x)\)

\(\displaystyle y''=(8A+6B)e^{3x}cos(x)+(8B-6A)e^{3x}sin(x)\)

You have \(\displaystyle C_{1}e^{x}+C_{2}e^{2x}\).

Now, do this problem like this one viewtopic.php?f=15&t=44678

You should get \(\displaystyle A=\frac{-3}{10}, \;\ B=\frac{1}{10}\)

 

[Jake89]

Sorry for a double thread, maybe an admin could delete the other one, something happened when I clicked submit, with my browser
thank you.