Answer to Question #311952 in Differential Equations for kxngToooch

Question

The equation of motion of a body is given by d²y/dt² + 4dy/dt + 13y = e^2tcost , where y is the distance and t is the time.

Determine a general solution for y in terms of t.

Answer 1

The given equation can be written as $(D^{2}+4D+13)y = e^{2t} \cos t$.

The auxiliary equation is $m^2+4m+13=0$ . Solving for m,

$\begin{aligned} m &= \frac{-4\pm\sqrt{16-52}}{2}\\ m &= \frac{-4\pm 6i}{2}\\ m & = -2\pm 3i \end{aligned}$

The complementary function is $\text{C.F} = e^{-2t}(c_1 \cos 3t + c_{2}\sin 3t)$

$\begin{aligned} \text{P.I} &= \dfrac{1}{D^2 + 4D+13} e^{2t} \cos t\\ &= e^{2t}\Bigg(\dfrac{1}{(D+2)^2 + 4(D+2) +13}\Bigg) \cos t\\ &= e^{2t}\Bigg(\dfrac{1}{D^2+4D+4 + 4D +8 +13}\Bigg) \cos t\\ &= e^{2t}\Bigg(\dfrac{1}{D^2+8D+25}\Bigg) \cos t\\ &= e^{2t}\Bigg(\dfrac{1}{-1+8D+25}\Bigg) \cos t\\ &= e^{2t}\Bigg(\dfrac{1}{24+8D}\Bigg) \cos t\\ &= e^{2t}\Bigg(\dfrac{24-8D}{24^2-64D^2}\Bigg) \cos t\\ &= e^{2t}\Bigg(\dfrac{24\cos t-8D(\cos t)}{576+64} \Bigg)\\ &= e^{2t}\Bigg(\dfrac{24\cos t+8\sin t}{640}\Bigg) \\ &= e^{2t}\Bigg(\dfrac{3\cos t+\sin t}{80}\Bigg) \\ \end{aligned}$

The general solution is $y = e^{-2t}(c_1 \cos 3t + c_{2}\sin 3t)+\dfrac{e^{2t}}{80}(3\cos t+\sin t)$