7.2.2 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.