11 Linear combination of two solution is solution to ODE

If \(y_{1},y_{2}\) are two solutions to \(ay^{\prime \prime }+by^{\prime }+cy=0\) then to show that \(c_{1}y_{1}+c_{2}y_{2}\) is also solution:

\begin{align*} ay_{1}^{\prime \prime }+by_{1}^{\prime }+cy_{1} & =0\\ ay_{2}^{\prime \prime }+by_{2}^{\prime }+cy_{2} & =0 \end{align*}

Multiply the first ODE by \(c_{1}\) and second ODE by \(c_{2}\)

\begin{align*} a\left ( c_{1}y_{1}\right ) ^{\prime \prime }+b\left ( c_{1}y_{1}\right ) ^{\prime }+c\left ( c_{1}y_{1}\right ) & =0\\ a\left ( c_{2}y_{2}\right ) ^{\prime \prime }+b\left ( c_{2}y_{2}\right ) ^{\prime }+c\left ( c_{2}y_{2}\right ) & =0 \end{align*}

Add the above two equations, using linearity of differentials

\[ a\left ( c_{1}y_{1}+c_{2}y_{2}\right ) ^{\prime \prime }+b\left ( c_{1}y_{1}+c_{2}y_{2}\right ) ^{\prime }+c\left ( c_{1}y_{1}+c_{2}y_{2}\right ) =0 \]

Therefore \(c_{1}y_{1}+c_{2}y_{2}\) satisfies the original ODE. Hence solution.