Question

Round three (-_-):

Answer 1

fight!

Answer 2

\[y'''+4y'=e^{x}\cos(2x).\]Lambda substitution:\[\lambda^3+4\lambda=0,\]\[\lambda(\lambda^2+4)=0.\]This has messy solutions:\[\lambda_{1}=0,\]\[\lambda_{2}=2i,\]\[\lambda_{3}=-2i.\]Therefore, its solution is:\[y=c_{1}\cos(2x)+c_{2}\sin(2x).\]The RHS is annihilated by\[[D^2-2D+5](D^3+4D)y=0,\]which is\[[\lambda^2-2\lambda+5](\lambda^3+4\lambda)=0,\]\[[\lambda^2-2\lambda+5]\lambda(\lambda^2+4)=0.\]This has roots:\[\lambda_{1}=0,\]\[\lambda_{2}=2i,\]\[\lambda_{3}=-2i,\]\[\lambda_{4}=1+2i,\]\[\lambda_{5}=1-2i.\]Therefore, we obtain the solution:\[y=c_{1}+c_{2}\cos(2x)+c_{3}\sin(2x)+e^{x}[c_{4}\cos(2x)+c_{5}\sin(2x)].\]Observe that\[y_{p}=A+e^{x}[B\cos(2x)+C\sin(2x)],\]\[y'_{p}=e^{x}[(C-2B)\sin(2x)+(B+2C)\cos(2x)],\]\[y''_{p}=-e^{x}[(4B+3C)\sin(2x)+(3B-4C)\cos(2x)],\]\[y'''_{p}=-e^{x}[(11C-2B)\sin(2x)+(11B+2C)\cos(2x)].\]Therefore,\[y'''+4y'=-e^{x}((6B+7C)\sin(2x)+(7B-6C)\cos(2x)),\]\[-e^{x}((6B+7C)\sin(2x)+(7B-6C)\cos(2x))=e^{x}\cos(2x),\]and\[A=0,\]\[-e^{x}((6B+7C)\sin(2x))=0,\]\[-e^{x}((7B-6C)\cos(2x))=e^{x}\cos(2x).\]This has solutions:\[A=0,\]\[B=-\frac{7}{85},\]\[C=\frac{6}{85}.\]Therefore, the final solution should be:\[y=e^{x}[-\frac{7}{85}\cos(2x)+\frac{6}{85}\sin(2x)]+c_{1}\cos(2x)+c_{2}\sin(2x).\]This one frustrated me so much...

Answer 3

I don't get it....

Did you need help with this or something?

Answer 4

Oh, yes, I forgot to ask for corroboration... -_-

Heh, I wouldn't be impressed if I made a mistake somewhere in there (I skipped a dozen or so steps).

Answer 5

So I guess the original question was to find y

Answer 6

Yep...

... I just noticed I'm missing a third constant as well as a couple 1/2's somewhere in there.

x_x

Answer 7

I'm very mistake prone. I can't do all of that and not make a mistake

Answer 8

The solution turned out to be okay. All I left out was the third c constant. Also, the quotients multiplying the c constants get absorbed. That's what confused me there for a bit.