Question

Find the second order linear differential equation whose general solution is given by y=C1 cos4t + C2 sin4t -e^t sint

Answer 1

The general solution of second order linear differential equation is
$$ y=C_{1} \cos 4 t+C_{2} \sin 4 t-e^{t} \sin t $$ We have to find second order linear differential equation whose solution is y. First differentiate (1), we have :
$$ y^{\prime}=-4 C_{1} \sin 4 t+4 C_{2} \cos 4 t-e^{t} \sin t-e^{t} \cos t $$ Again differentiate, we get :
$$ y^{\prime \prime}=-16 C_{1} \cos 4 t-16 C_{2} \sin 4 t+e^{t} \sin t-e^{t} \cos t-e^{t} \cos t-e^{t} \sin t $$ $=-16 C_{1} \cos 4 t-16 C_{2} \sin 4 t-2 e^{t} \cos t$
Now we multiply (1) by 16 and add with (2), we have :
$y^{\prime \prime}+16 y=-16 C_{1} \cos 4 t-16 C_{2} \sin 4 t-2 e^{t} \cos t+16 C_{1} \cos 4 t+16 C_{2} \sin 4 t-$
$16 e^{t} \sin t$
$y^{\prime \prime}+16 y=-2 e^{t} \cos t-16 e^{t} \sin t$
$=-2 e^{t}(\cos t+8 \sin t)$
HENCE THE REOUIRED SECOND ORDER LINEAR DIFFERENTIAL EGUATION IS:
$y^{\prime \prime}+16 y=-2 e^{t}(\cos t+8 \sin t)$