Question

What is x(t)?
x''+ax'+bx=c+cos(dt+p)

Answer 1

hooowahhhhhh
man i dont want try this without computing engine

http://www.wolframalpha.com/input/?i=y%27%27%28x%29%2Ba*y%27%28x%29%2Bb*y%28x%29%3Dc%2Bcos%28d*x%2Bp%29

Answer 2

LOL

Answer 3

\[x_h = ae^{r_1t} + be^{r_2t}\]
\[r_1 = \frac{-a +\sqrt{a^2-4b}}{2}\]\[r_2 = \frac{-a -\sqrt{a^2-4b}}{2}\]
\[b*x_p = bAe^{r_1t} + bBe^{r_2t};A(t),B(t)\]
\[a*x'_p = (A'e^{r_1t} + B'e^{r_2t}=0)\]\[\hspace{3em} +aAr_1e^{r_1t} + aBr_2e^{r_2t}\]
\[x''_p = A'r_1e^{r_1t} + B'r_2e^{r_2t}\]\[\hspace{3em} +Ar^2_1e^{r_1t} + Br^2_2e^{r_2t}\]
\[c+cos(u) = bAe^{r_1t} + bBe^{r_2t}+ aAr_1e^{r_1t} + aBr_2e^{r_2t}+ A'r_1e^{r_1t} + B'r_2e^{r_2t}\]\[\hspace{8em} +Ar^2_1e^{r_1t} + Br^2_2e^{r_2t}\]
\[c+cos(u) = Ae^{r_1t}(b+ar_1+r^2_1) + Be^{r_2t}(b+ar_2+r^2_2)+ A'r_1e^{r_1t} + B'r_2e^{r_2t}\]
ideally: \(Ae^{r_1t}(b+ar_1+r^2_1) + Be^{r_2t}(b+ar_2+r^2_2)=0\)
to give us

\[0=A'e^{r_1t} + B'e^{r_2t}\]\[c+cos(u) = A'r_1e^{r_1t} + B'r_2e^{r_2t}\]which is a system of 2 equations in 2 unknowns