Missing \(y\) as in \(ay^{\prime \prime \prime }+by^{\prime \prime }+cy^{\prime }=f\left ( x\right ) \)

ode internal name "higher_order_ODE_missing_y"

Since \(y\) is missing, we then assume \(y^{\prime }=u\) and the ode reduces to one order less. Now the lower order ode is solved.

Example 1

\[ x^{2}y^{\prime \prime \prime }+xy^{\prime \prime }+y^{\prime }=0 \]

This is not Euler type as it stands. Let \(y^{\prime }=u\) then the ode becomes

\[ x^{2}u^{\prime \prime }+xu^{\prime }+u=0 \]

This is now Euler type. Solving it gives

\[ u=c_{2}\cos \left ( \ln x\right ) +c_{3}\sin \left ( \ln x\right ) \]

Hence

\[ y^{\prime }=c_{2}\cos \left ( \ln x\right ) +c_{3}\sin \left ( \ln x\right ) \]

Solving this as first order ode of quadrature type gives

\begin{align*} y & =\frac {c_{2}}{2}x\cos \left ( \ln x\right ) +\frac {c_{2}}{2}x\sin \left ( \ln x\right ) -\frac {1}{2}c_{3}x\cos \left ( \ln x\right ) +\frac {1}{2}c_{3}x\sin \left ( \ln x\right ) +c_{1}\\ & =x\cos \left ( \ln x\right ) \left ( \frac {c_{2}}{2}-\frac {1}{2}c_{3}\right ) +x\sin \left ( \ln x\right ) \left ( \frac {c_{2}}{2}++\frac {1}{2}c_{3}\right ) +c_{1}\\ & =C_{2}x\cos \left ( \ln x\right ) +C_{3}x\sin \left ( \ln x\right ) +c \end{align*}

Example 2

\[ xy^{\prime \prime \prime \prime }+y^{\prime \prime \prime }+y^{\prime \prime }=0 \]

Let \(y^{\prime }=u\) then the ode becomes

\[ xu^{\prime \prime \prime }+u^{\prime \prime }+u^{\prime }=0 \]

Since \(u\) is missing then let \(u^{\prime }=v\) and the above becomes

\[ xv^{\prime \prime }+v^{\prime }+v=0 \]

This is now second order ode. This is Bessel ode whose solution is

\[ v=c_{3}\operatorname {BesselJ}_{0}\left ( 2\sqrt {x}\right ) +c_{4}\operatorname {BesselY}_{0}\left ( 2\sqrt {x}\right ) \]

Hence

\[ u^{\prime }=c_{3}\operatorname {BesselJ}_{0}\left ( 2\sqrt {x}\right ) +c_{4}\operatorname {BesselY}_{0}\left ( 2\sqrt {x}\right ) \]

This is solved by quadrature giving

\[ u=c_{3}\sqrt {x}\operatorname {BesselJ}_{1}\left ( 2\sqrt {x}\right ) +c_{4}\sqrt {x}\operatorname {BesselY}_{1}\left ( 2\sqrt {x}\right ) +c_{2}\]

Hence

\[ y^{\prime }=c_{3}\sqrt {x}\operatorname {BesselJ}_{1}\left ( 2\sqrt {x}\right ) +c_{4}\sqrt {x}\operatorname {BesselY}_{1}\left ( 2\sqrt {x}\right ) +c_{2}\]

This is solved by quadrature giving

\[ y=c_{3}x\operatorname {BesselJ}_{2}\left ( 2\sqrt {x}\right ) +c_{4}x\operatorname {BesselY}_{2}\left ( 2\sqrt {x}\right ) +c_{2}x+c_{1}\]

Example 3

\[ xy^{\prime \prime \prime }-y^{\prime \prime }=0 \]

Let \(y^{\prime }=u\) then the ode becomes

\[ xu^{\prime \prime }-u^{\prime }=0 \]

Since \(u\) is missing then let \(u^{\prime }=v\) and the above becomes

\[ xv^{\prime }-v=0 \]

This is linear first order ode whose solution is \(v=c_{1}x\). Hence \(u^{\prime }=c_{1}x\). Integrating gives \(u=c_{1}x^{2}+c_{2}\). Hence

\[ y^{\prime }=c_{1}x^{2}+c_{2}\]

Integrating gives

\[ y=c_{1}x^{3}+c_{2}x+c_{3}\]