Question

Round four (>:P)

Answer 1

\[y^{(4)}+8y'=4.\]Lambda substitution:\[\lambda^4+8\lambda=0,\]\[\lambda(\lambda^3+8),\]\[\lambda(\lambda + 2)(\lambda^2-2\lambda+4).\]This has solutions:\[\lambda_{1}=0,\]\[\lambda_{2}=-2,\]\[\lambda_{3}=1+2\sqrt{3}i,\]\[\lambda_{4}=1-2\sqrt{3}i.\]Therefore, the general solution should be:\[y=c_{1}+c_{2}e^{-2x}+e^{x}[c_{3}\cos(2\sqrt{3}x)+c_{4}\sin(2\sqrt{3}x)].\]The RHS is eliminated by:\[D(D^4+8D)y=0,\]\[D^2(D+2)(D^2-2D+4)y=0\implies\]\[\lambda^2(\lambda+2)(\lambda^2-2\lambda+4)=0.\]Solutions:\[\lambda_{1}=\lambda_{2}=0,\]\[\lambda_{3}=-2,\]\[\lambda_{4}=1+2\sqrt{3}i,\]\[\lambda_{5}=1-2\sqrt{3}i.\]Then,\[y=c_{1}+c_{2}x+c_{3}e^{-2x}+e^{x}[c_{4}\cos(2\sqrt{3}x)+c_{5}\sin(2\sqrt{3}x)].\]This implies:\[y_{p}=Ax,\]\[y'_{p}=A,\]\[y''_{p}=y'''_{p}=y^{(4)}_{p}=0.\]Thus,\[y^{(4)}_{p}+8y'_{p}=8A=4,\]and\[A=\frac{1}{2}.\]Finally,\[y=c_{1}+c_{2}e^{-2x}+e^{x}[c_{3}\cos(2\sqrt{3}x)+c_{4}\sin(2\sqrt{3}x)+\frac{1}{2}x.\]I know I could've cut a few steps here and there (long way), but does this seem like a legit solution?

спасибо

Answer 2

Yes.

But for what it's worth, I still think the method we discussed last night (and in the MIT lecture!) is a more robust way to obtain the particular solution.

Answer 3

(In this case, you need the more general version discussed in the lecture where the formal polynominal of the differential operator p(D) is equal to zero for the input exponential.)