1.3.3.3 \(y^{\prime }+y=\sin \left ( x\right ) \) with two IC \(y\left ( 0\right ) =0,y^{\prime }\left ( 0\right ) =0\)

The general solution is \(y=-\frac {\cos x}{2}+\frac {\sin \left ( x\right ) }{2}+c_{1}e^{-x}\). Since we have two IC’s and only one constant, then this is an overdetermined case. One equation is generated for each IC. This gives

\begin{align*} y\left ( 0\right ) & =-\frac {\cos \left ( 0\right ) }{2}+\frac {\sin \left ( 0\right ) }{2}+c_{1}e^{-\left ( 0\right ) }\\ 0 & =-\frac {1}{2}+c_{1}\end{align*}

For the second IC, we first differentiate the ode

\[ y^{\prime }\left ( x\right ) =\frac {\sin x}{2}+\frac {\cos \left ( x\right ) }{2}-c_{1}e^{-x}\]

Applying the second IC the above becomes

\begin{align*} y^{\prime }\left ( 0\right ) & =\frac {\sin \left ( 0\right ) }{2}+\frac {\cos \left ( 0\right ) }{2}-c_{1}e^{-\left ( 0\right ) }\\ 0 & =\frac {1}{2}-c_{1}\end{align*}

The two equations we ended up with are the following

\begin{align*} 0 & =-\frac {1}{2}+c_{1}\\ 0 & =\frac {1}{2}-c_{1}\end{align*}

We see it is the same equation. Hence the solution is \(c_{1}=\frac {1}{2}\).