3.8.1.1 Example 1

Solve

\[ y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=0 \]

Given that one solution  is known to be \(y_{1}\). We start by assuming the second solution is \(y_{2}=y_{1}u\left ( x\right ) \) where \(u\left ( x\right ) \) is to be determined. Hence

\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 }\end{align*}

Substituting in the given ODE gives (since \(y_{2}\) is a solution, then it also satisfies the ode)

\[ \left ( y_{1}^{\prime \prime }u+2y_{1}^{\prime }u^{\prime }+y_{1}u^{\prime \prime }\right ) +p\left ( y_{1}^{\prime }u+y_{1}u^{\prime }\right ) +qy_{1}u=0 \]

And now we collect on \(u\) and all its derivatives. The above becomes

\[ u\left ( y_{1}^{\prime \prime }+py_{1}^{\prime }+qy_{1}\right ) +u^{\prime }\left ( 2y_{1}^{\prime }+py_{1}\right ) +y_{1}u^{\prime \prime }=0 \]

But \(y_{1}^{\prime \prime }+py_{1}^{\prime }+qy_{1}=0\). The above becomes

\[ u^{\prime }\left ( 2y_{1}^{\prime }+py_{1}\right ) +y_{1}u^{\prime \prime }=0 \]

Ok, you migth ask, what did we accomplish in all of this? Since we eneded up with just another second order ode. But here is the main point of this method. This new ode is missing the \(u\) term. Therefore by letting \(u^{\prime }=v\) we can make the above ode become first order ode

\[ v\left ( 2y_{1}^{\prime }+py_{1}\right ) +y_{1}v^{\prime }=0 \]

SInce \(y_{1}\) is given, the above first order ode is now solved for \(v\), and once \(v\) is known, then \(u\) is found by integrating \(u^{\prime }=v\) and once \(u\) is found then \(y_{2}\) is found from \(y_{2}=y_{1}u\left ( x\right ) \).

The above ode can be written as

\[ v^{\prime }+\left ( 2\frac {y_{1}^{\prime }}{y_{1}}+p\right ) v=0 \]

Hence it is linear first order ode. The integrating factor is

\begin{align*} \mu & =e^{\int 2\frac {y_{1}^{\prime }}{y_{1}}+pdx}\\ & =e^{\int \frac {2}{y_{1}}\frac {dy_{1}}{dx}dx+\int pdx}\\ & =e^{\int \frac {2}{y_{1}}dy_{1}+\int pdx}\\ & =e^{2\ln y_{1}+\int pdx}\\ & =e^{2\ln y_{1}}e^{\int pdx}\\ & =y_{1}^{2}e^{\int pdx}\end{align*}

Therefore

\begin{align} d\left ( v\mu \right ) & =0\nonumber \\ v\mu & =c_{1}\nonumber \\ v & =c_{1}\frac {e^{-\int pdx}}{y_{1}^{2}} \tag {1}\end{align}

Since \(u^{\prime }=v\) then we have

\[ \frac {du}{dx}=v \]

Integrating

\[ u=\int vdx+c_{2}\]

Here we are free to let \(c_{2}=0\). Therefore

\begin{equation} u=\int vdx \tag {2}\end{equation}

Therefore

\begin{align} y_{2} & =y_{1}u\nonumber \\ & =y_{1}\int vdx\nonumber \\ & =y_{1}\int \left ( c_{1}\frac {e^{-\int pdx}}{y_{1}^{2}}\right ) dx\nonumber \\ & =c_{1}y_{1}\int \left ( \frac {e^{-\int pdx}}{y_{1}^{2}}\right ) dx \tag {3}\end{align}

And the solution is

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

The following example shows how the above can be applied to a concrete problem.