Question

The question is too long. Thus,

Answer 1

Someone help me corroborate:\[y''-5y=x^2-2x.\]This becomes:\[\lambda^2-5=0,\]\[\lambda^2=5,\]\[\lambda=\pm\sqrt{5}.\]Thus,\[y=c_{1}e^{\sqrt{5}x}+c_{2}e^{-\sqrt{5}x}.\]RHS can be anihilated by:\[D^3(D^2-5)y=0.\]Thus\[\lambda^3(\lambda^2-5)=0\]has the solutions:\[\lambda_{1}=\lambda_{2}=\lambda_{3}=0,\]\[\lambda_{4}=\sqrt{5},\]\[\lambda_{6}=-\sqrt{5},\]which gives:\[c_{1}+c_{2}x+c_{3}x^2+c_{4}e^{\sqrt{5}x}+c_{5}e^{-\sqrt{5}x}.\]We then see that\[y_{p}=A+Bx+Cx^2,\]\[y'_{p}=B+2Cx,\]\[y''_{p}=2C.\]Therefore\[y''_{p}-5y=2C-5A-5Bx-5Cx^2=x^2-2x.\]Solving for A, B, and C, we obtain:\[2C-5A=0,\]\[-5B=2,\]\[-5C=1,\]and\[C=-\frac{1}{5},\]\[B=-\frac{2}{5},\]\[A=-\frac{2}{25}.\]Therefore, my solution should be:\[y=-\frac{2}{25}-\frac{2}{5}x-\frac{1}{5}x^2+c_{1}e^{\sqrt{5}x}+c_{2}e^{-\sqrt{5}x}.\]Did I make a mistake anywhere?

Answer 2

\[\text{Where's }\lambda_5?\]

Answer 3

Sorry, up there I meant
\[\lambda_{5}\]not\[\lambda_{6}.\]

Answer 4

should be +2/5x instead of -2/5x

Answer 5

Oh yes! Excellent observation. Thank you very much! :D

Does the rest look fine, btw?

Answer 6

Yes.

Answer 7

Danke schön. :3

Answer 8

Bitte