Solved using reduction of order

ode internal name "higher_order_reduction_of_order"

Given third order ode, which is linear (this method actually works for constant or non-constant coefficients), such as

\[ y^{\prime \prime \prime }+ay^{\prime \prime }+by^{\prime }+cy=0 \]

And given one known solution, \(y_{1}\left ( x\right ) \) then let second solution be \(y_{2}=y_{1}u\) (there will be three independent basis solutions, since this is third order ode). Then substituting this into the ode gives

\begin{align*} y_{2}^{\prime } & =y_{1}^{\prime }u+y_{1}u^{\prime }\\ y_{2}^{\prime \prime } & =y_{1}^{\prime \prime }u+y_{1}^{\prime }u^{\prime }+y_{1}^{\prime }u^{\prime }+y_{1}u^{\prime \prime }\\ & =y_{1}^{\prime \prime }u+2y_{1}^{\prime }u^{\prime }+y_{1}u^{\prime \prime }\\ y_{2}^{\prime \prime \prime } & =y_{1}^{\prime \prime \prime }u+y_{1}^{\prime \prime }u^{\prime }+2y_{1}^{\prime \prime }u^{\prime }+2y_{1}^{\prime }u^{\prime \prime }+y_{1}^{\prime }u^{\prime \prime }+y_{1}u^{\prime \prime \prime }\\ & =y_{1}^{\prime \prime \prime }u+3y_{1}^{\prime \prime }u^{\prime }+3y_{1}^{\prime }u^{\prime \prime }+y_{1}u^{\prime \prime \prime }\end{align*}

Substituting the above into the given original ode (since \(y_{2}\) is a solution, then it satisfies the ode), gives

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

But \(y_{1}^{\prime \prime \prime }+ay_{1}^{\prime \prime }+by_{1}^{\prime }+cy_{1}=0\). The above becomes

\begin{equation} 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 \tag {1}\end{equation}

Since there is no \(u\) term, then let \(v=u^{\prime }\) and the above reduces to second order ode

\begin{equation} v\left ( 3y_{1}^{\prime \prime }+2ay_{1}^{\prime }+by_{1}\right ) +v^{\prime }\left ( 3y_{1}^{\prime }+ay_{1}\right ) +v^{\prime \prime }y_{1}=0 \tag {2}\end{equation}

Solving for \(v\) from the above, then we find \(u\) since \(u^{\prime }=v\), by integration, we introduces one more constant of integration which we can set to zero. Once we find \(u\) then we can find the second solution \(y_{2}\) since \(y_{2}=y_{1}u.\) Then the final solution is

\[ y=c_{1}y_{1}+c_{2}y_{2}\]

Note that \(y_{2}\) which was found above, comes with 2 basis solutions in it. So the above gives the three basis solutions needed.

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.

Example 2

\begin{align*} y^{\prime \prime \prime }-\frac {2}{3}y^{\prime \prime }+4y^{\prime }-\frac {8}{3}y & =0\\ y_{1} & =e^{\frac {2x}{3}}\end{align*}

Let \(y_{2}=y_{1}u=ue^{\frac {3}{2}x}\). Where here \(a=-\frac {2}{3},b=4,c=-\frac {8}{3}.\) 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 }+\left ( 2\right ) \left ( -\frac {2}{3}\right ) y_{1}^{\prime }+4y_{1}\right ) +u^{\prime \prime }\left ( 3y_{1}^{\prime }-\frac {2}{3}y_{1}\right ) +u^{\prime \prime \prime }y_{1} & =0\\ u^{\prime }\left ( 3y_{1}^{\prime \prime }-\frac {4}{3}y_{1}^{\prime }+4y_{1}\right ) +u^{\prime \prime }\left ( 3y_{1}^{\prime }-\frac {2}{3}y_{1}\right ) +u^{\prime \prime \prime }y_{1} & =0 \end{align*}

But \(y_{1}=e^{\frac {2x}{3}},y_{1}^{\prime }=\frac {2}{3}e^{\frac {2x}{3}},y_{1}^{\prime \prime }=\frac {4}{9}e^{\frac {2x}{3}}\). Hence the above becomes

\begin{align*} u^{\prime }\left ( 3\left ( \frac {4}{9}e^{\frac {2x}{3}}\right ) -\frac {4}{3}\left ( \frac {2}{3}e^{\frac {2x}{3}}\right ) +4\left ( e^{\frac {2x}{3}}\right ) \right ) +u^{\prime \prime }\left ( 3\left ( \frac {2}{3}e^{\frac {2x}{3}}\right ) -\frac {2}{3}\left ( e^{\frac {2x}{3}}\right ) \right ) +u^{\prime \prime \prime }e^{\frac {2x}{3}} & =0\\ u^{\prime }\left ( \frac {4}{3}-\frac {8}{9}+4\right ) +u^{\prime \prime }\left ( 2-\frac {2}{3}\right ) +u^{\prime \prime \prime } & =0\\ \frac {40}{9}u^{\prime }+\frac {4}{3}u^{\prime \prime }+u^{\prime \prime \prime } & =0\\ 40u^{\prime }+12u^{\prime \prime }+9u^{\prime \prime \prime } & =0 \end{align*}

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

\[ 40v+12v^{\prime }+9v^{\prime \prime }=0 \]

This is now second order ode. The solution for \(v\) can be found to be

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

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

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

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

\begin{align*} u & =\int c_{1}e^{\frac {-2x}{3}}\sin \left ( 2x\right ) +c_{2}e^{-\frac {2x}{3}}\cos \left ( 2x\right ) dx\\ & =-\frac {9e^{-\frac {2x}{3}}}{20}\left ( \left ( c_{1}+\frac {c_{2}}{3}\right ) \cos \left ( 2x\right ) +\frac {1}{3}\left ( c_{1}-3c_{2}\right ) \sin \left ( 2x\right ) \right ) \\ & =e^{-\frac {2x}{3}}\left ( c_{3}\cos \left ( 2x\right ) +c_{4}\sin \left ( 2x\right ) \right ) \end{align*}

Where in the last step above, constants were combined to make new constants. Renaming constants, the above becomes

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

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

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

Hence the solution is

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

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