Please take a look at my work below and provide a good critique:
Solve the differential equation using the method of undetermined coefficients or variation of parameters.
y'' - 3y' + 2y = sin(x)
yc(x)= c1*e^2x+c2*e^x
y"-3y'+2y=sin(x)
r^2-3r+2=0
(r-1)(r-2)=0
r1=1, r2=2
yp(x)=Acosx+Bsinx
yp'(x)=-Asinx+Bcosx
yp"(x)=-Acosx-Bsinx
(-Acosx-Bsinx)-3(-Asinx+Bcosx)+2(Acosx+Bsinx)=sinx
-Acosx-Bsinx+3Asinx-3Bcosx+2Acosx+2Bsinx
(A-3B)cosx+(B+3A)sinx=sinx
A-3B=0 and B+3A=1
A=3B, B+3B=1
A=3/4, B=1/4
yp(x)=(3/4)cosx+(1/4)sinx=(1/4)(3cosx+sinx)
This did not come out correctly. I am trying to solve for yp(x). Where have I made an error and how can I fix it? Thanks.
-Acosx-Bsinx+3Asinx-3Bcosx+2Acosx+2Bsinx=sinx
from that line, to the next,
(A-3B)cosx+(B+3A)sinx=sinx
I think you should have
(A-3B+2A)cosx+(B+3A)sinx=sinx
check my adding.
got it, thanks.
1 answer