7.2.3.1 Example 1

\begin{align*} y^{\prime \prime \prime }+3y^{\prime \prime }-54y & =0\\ y_{1} & =e^{3x}\end{align*}

Let \(y_{2}=ue^{3x}\). Where here \(a=3,b=0,c=-54.\) Then EQ (1) becomes

\begin{align*} u^{\prime }\left ( 3y_{1}^{\prime \prime }+2ay_{1}^{\prime }+by_{1}\right ) +u^{\prime \prime }\left ( 3y_{1}^{\prime }+ay_{1}\right ) +u^{\prime \prime \prime }y_{1} & =0\\ u^{\prime }\left ( 3y_{1}^{\prime \prime }+6y_{1}^{\prime }\right ) +u^{\prime \prime }\left ( 3y_{1}^{\prime }+3y_{1}\right ) +u^{\prime \prime \prime }y_{1} & =0 \end{align*}

But \(y_{1}=e^{3x},y_{1}^{\prime }=3e^{3x},y_{1}^{\prime \prime }=9e^{3x}\). Hence the above becomes

\begin{align*} u^{\prime }\left ( 27e^{3x}+18e^{3x}\right ) +u^{\prime \prime }\left ( 9e^{3x}+3e^{3x}\right ) +u^{\prime \prime \prime }e^{3x} & =0\\ 45u^{\prime }+12u^{\prime \prime }+u^{\prime \prime \prime } & =0 \end{align*}

Let \(u^{\prime }=v\,\) then the above becomes

\[ 45v+12v^{\prime }+v^{\prime \prime }=0 \]

This is now second order ode. The solution for \(v\) is

\[ v=c_{1}e^{-6x}\sin \left ( 3x\right ) +c_{2}e^{-6x}\cos \left ( 3x\right ) \]

But \(u^{\prime }=v\), then

\[ u=\int vdx+c_{3}\]

We can choose \(c_{3}=0\,\). Hence

\begin{align*} u & =\int \left ( c_{1}e^{-6x}\sin \left ( 3x\right ) +c_{2}e^{-6x}\cos \left ( 3x\right ) \right ) dx\\ & =\frac {e^{-6x}}{15}\left ( \left ( c_{1}+2c_{2}\right ) \cos \left ( 3x\right ) +2\left ( c_{1}-\frac {c_{2}}{2}\right ) \sin \left ( 3x\right ) \right ) \\ & =e^{-6x}\left ( c_{3}\cos \left ( 3x\right ) +c_{4}\sin \left ( 3x\right ) \right ) \end{align*}

Where in the last step above, we merged constants to make new constants. Renaming constants back gives

\[ u=e^{-6x}\left ( c_{1}\cos \left ( 3x\right ) +c_{2}\sin \left ( 3x\right ) \right ) \]

Hence since second solution is \(y_{2}=y_{1}u\) then we have

\begin{align*} y_{2} & =y_{1}u\\ & =e^{3x}\left ( e^{-6x}\left ( c_{1}\cos \left ( 3x\right ) +c_{2}\sin \left ( 3x\right ) \right ) \right ) \\ & =e^{-3x}\left ( c_{1}\cos \left ( 3x\right ) +c_{2}\sin \left ( 3x\right ) \right ) \\ & =c_{1}e^{-3x}\cos \left ( 3x\right ) +c_{2}e^{-3x}\sin \left ( 3x\right ) \end{align*}

Hence the solution is

\begin{align*} y & =c_{3}y_{1}+c_{4}y_{2}\\ & =c_{3}e^{3x}+c_{4}\left ( c_{1}e^{-3x}\cos \left ( 3x\right ) +c_{2}e^{-3x}\sin \left ( 3x\right ) \right ) \\ & =c_{3}e^{3x}+c_{1}e^{-3x}\cos \left ( 3x\right ) +c_{2}e^{-3x}\sin \left ( 3x\right ) \end{align*}

Where in the last step above we just merged and renamed constants.