Question

2nd order ODE

Answer 1

Solve \(y''+y'-2y=sin^2x\)

Answer 2

\(y''+y'-2y=\frac{1}{2}-\frac{1}{2}cos~2x\)

Answer 3

\(m^2+m-2=0\)

\((m-1)(m+2)=0\)

\(m=1,m=-2\)

\(y_h(x)=Ae^x+Be^{-2x}\)

\(y_p=y_{p_1}+y_{p_2}\)

\(y_{p_1}=C_0\)

\(y'_{p_1}=0\)

\(y''_{p_2}=0\)

\(y_{p_1}=C_o=\frac{1}{2}-\frac{1}{2}cos~2x\)

\(y_{p_2}=Acos2x+Bsin2x\)

\(y'_{p_2}=-2Asin2x+2Bcos2x\)

\(y''_{p_2}=-4Acos2x-4Bsin2x\)

\(-4Acos2x-4Bsin2x-2Asin2x+2Bcos2x-2(Acos2x+Bsin2x)=\frac{1}{2}-\frac{1}{2}cos2x\)

\((-4A+2B-2A)cos2x+(-4A-2A-2B)sin2x=\frac{1}{2}-\frac{1}{2}cos2x\)

\(-6A+2B=-\frac{1}{2}\)----(1)
\(-6B-2A=0\)----(2)

\(A=\frac{3}{40},B=-\frac{1}{40}\)

\(y_{p_2}=\frac{3}{40}cos2x-\frac{1}{40}sin2x\)

\(y(x)=y_h(x)+y_{p_1}+y_{p_2}\)

\(y(x)=Ae^x+Be^{-2x}+\frac{1}{2}-\frac{1}{2}cos2x+\frac{3}{40}cos2x-\frac{1}{40}sin2x\)

Answer 4

is it correct?

Answer 5

@nuts

Answer 6

Your costants A and B from the long drawn out sine and cosine business, yes those look correct.

I think you plugged in your first particular form incorrectly though.

\[\large\rm y''+y'-2y=\frac12-\frac12\cos2x\]Plugging in the yp1,\[\large\rm 0+0-2C_o=\frac12\]Which should give you \(\large\rm C_o=-1\), yes?

That's the only minor error I can find.

Answer 7

Woops, I forgot how to do these :P
I guess that second equation I wrote should be,\[\large\rm 0+0-2C_o=\frac12-\frac12\cos2x\]So then,\[\large\rm C_o=\cos2x-1\]

Answer 8

I dunno, one of those two should be correct XD

Answer 9

yeah,that's right X'D

Owh,I managed to solve it

I shouldn't hv derive \(y_{p_1}~~~and~~~y_{p_2}\) first

Answer 10

I should add both of them,then only derive \(y_p\)

Answer 11

Thanks 4 the help @Zepdrix ^~^