1.3.3.2 \(y^{\prime }+y=\sin \left ( x\right ) \) with one mixed IC \(y\left ( 0\right ) +y\left ( 1\right ) =0\)

The general solution is \(y=-\frac {\cos x}{2}+\frac {\sin \left ( x\right ) }{2}+c_{1}e^{-x}\) and since we have one IC given, then we set up one equation, this is done by evaluating each term in the IC which 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 ) }\\ & =-\frac {1}{2}+c_{1}\end{align*}

And

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

Then the IC condition becomes

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

We have one equation and one unknown. Solving the above for \(c_{1}\) gives the particular solution.