Answer to Question #206918 in Differential Equations for Nshalati

Question

y"-4y'+13y=sin2t

Answer 1

Solution.

y"-4y'+13y=sin2t

$\lambda^2-4\lambda+13=0$

$D=16-4•13=16-52=-36$

$\lambda=\frac{4\pm 6i}{2}=2\pm 3i$

From here solution of homogeneous equation $y''-4y'+13=0$ is

$y=C_1e^{2t}\sin{3t}+C_2e^{2t}\cos{3t},$

where $C_1,C_2$ are some constants.

Find particular solution of equation $y''-4y'+13=\sin{2t}$ in the form $y=A\sin{2t}+B\cos{2t}.$

Then

$y'=2A\cos{2t}-2B\sin{2t},$ and

$y''=-4A\sin{2t}-4B\cos{2t}.$

$(-4A+8B+13A)\sin{2t}+(-4B-8A+13B)\cos{2t}=\newline\sin{2t}$

Therefore, $A=\frac{9}{145},B=\frac{8}{145}.$

So, we have solution

$y=C_1e^{2t}\sin{3t}+C_2e^{2t}\cos{3t}+\frac{9}{145}\sin{2t}+\frac{8}{145}\cos{2t}.$

Answer. $y=C_1e^{2t}\sin{3t}+C_2e^{2t}\cos{3t}+\frac{9}{145}\sin{2t}+\frac{8}{145}\cos{2t}.$