Answer to Question #152992 in Differential Equations for Ajiesh

Question

solve the equation d²y/dt² + 2dy/dt -3y = sin t , given that y = dy/dt = 0 when t = 0

Answer 1

$\frac{d^2y}{dt^2}+2\frac{dy}{dt}-3y=\sin t$

The corresponding homogeneous equation $y''+2y'-3y=0$ has the characteristics equation $m^2+2m-3=0$

$(m-1)(m+3)=0\\ m=1\text{ or } m=-3$

$Y_c=C_1e^{1.t}+C_2e^{-3t}\\ Y_c=C_1e^t+C_2e^{-3t}$

The non-homogeneous term is $g(t)=\sin t$ . We choose $Y_p=A\cos t+ B\sin t$

$Y'_p=B\cos t-A\sin t, Y''_p=-(A\cos t+B\sin t)$ . Put these into the equation. We have;

$-(A\cos t+B\sin t)+2(B\cos t-A\sin t)-3(A\cos t+B\sin t)=\sin t\\ (-4A+2B)\cos t+(-2A-4B)\sin t=\sin t$

Compare both sides

$-4A+2B=0.................(1)\\ -2A-4B=1................(2)$

From (1), $A=\frac{B}{2}.$ Put this into (2)

$-5B=1,B=-\frac{1}{5}\\ A=-\frac{1}{10}$

Therefore,$Y_p=-\frac{1}{10}\cos t-\frac{1}{5}\sin t$

The general solution $y=Y_c+Y_p$

$y=C_1e^t+C_2e^{-3t}-\frac{1}{10}\cos t-\frac{1}{5}\sin t$

But, $y=y'=0$ when $t=0$ . This implies that,

$0=C_1e^0+C_2e^{-3.0}-\frac{1}{10}\cos 0-\frac{1}{5}\sin 0\\ 0=C_1+C_2-\frac{1}{10}\\ \frac{1}{10}=C_1+C_2....................(3)$

$y'=C_1e^t-2C_3e^{-3t}+\frac{1}{10}\sin t-\frac{1}{5}\cos t\\ 0=C_1e^0-3C_2e^{-3.0}+\frac{1}{10}\sin 0-\frac{1}{5}\sin 0\\ 0=C_1-3C_2-\frac{1}{5}\\ C_1-3C_2=\frac{1}{5}....................(4)$

Subtract (4) from (3)

$4C_2=\frac{1}{10}-\frac{1}{5}=-\frac{1}{10},C_2=-\frac{1}{40}$

Put this into (3)

$C_1=\frac{1}{10}+\frac{1}{40}=\frac{1}{8}$

Hence, $y=\frac{1}{8}e^t-\frac{1}{40}e^{-3t}-\frac{1}{10}\cos t-\frac{1}{5}\sin t$