Solution of a nonhomogeneous equation

[median27]

(i d0nt kn0w how to use LaTeX)

1)D^2(D-1)y=3e^x+sinx

for yc:
let y=e^mx
D^2(D-1)e^mx=0
m^2(m-1)e^mx=0
f(m)=0
m^2(m-1)=0
m=0,0,1

yc= C1+C2x+C3e^x

for yp:
R(x)=3e^x+sinx
m'=1,+/- 1i

yp=Axe^x+Bcosx+Csinx
yp'=A(xe^x+e^x)-Bsinx+Ccosx
yp"=A(xe^x+2e^x)-Bcosx-Csinx

D^2(D-1)yp=3e^x+sinx

Guys can you help me find the value of A, B and C. I had a hard time simplifying the terms and can't equate the coefficients. Thanks for analyzing my post this far. :D

 

[vela]

You're doing fine so far. You need to differentiate one more time because

D²(D-1)y_p = (D³-D²)y_p=y_p'''-y_p''

 

[median27]

Thanks for your help vela! I simply overlooked the problem and didn't get the third derivative. :D

 

[HallsofIvy]

Here's how I would have done it:
If $D^@(D-1)y= 3e^x+ sin(x)$ then $D(D-1)y= 3e^x- cos(x)+ C$ and $(D-1)y= 3e^x- sin(x)+ Cx+ D$

Then solve the characteristic equation m- 1= 0 to get m= 1 which means that $e^x$ is a solution to the associated homogeneous equation. Since part of the right side is $e^x$ look for a solution of the form $Axe^x+ Bsin(x)+ Ccos(x)+ Ex+ F$.

 

[median27]

Can you figure out my mistake. The answer is supposed to be:
y=C1e^-x+C2e^-2x+6x^2-18x+21

Here is the problem and my solution:

2) (D^2+3D+2)y=12x^2

for yc
let y=e^mx
(D^2+3D+2)e^mx=0
(m^2+3m+2)e^mx=0
f(m)=0
m^2+3m+2=0
m=-1,-2

yc=C1e^-x+C2e^-2x

for yp
R(x)=12x^2
m'=0,0,0
yp=A+Bx+Cx^2
yp'=B+2Cx^2
yp"=4Cx^2

(D^2+3D+2)yp=12x^2

4Cx^2+3B+6Cx^2+2A+2Bx+2Cx^2=12x^2

3B+12Cx^2+2A+2Bx=12x^2

Equating coefficients
i get C=1, B&A=0

Giving:
yp=x^2

Answer
y=C1e^-x+C2e^-2x+x^2
(which is wr0ng)

Thanks for any help!

 

[median27]

P.s. An0ther problem was posted.

 

[vela]

You didn't calculate the derivatives of y_p correctly.

 

[median27]

Alright, i get it. :D I'm used to differentiate e^n terms and overlooked x^n terms. I'm becoming reckless. Thanks vela.