3.2.1.1 Example 1 (Variation of parameters)

\[ 4y^{\prime \prime }-y=e^{\frac {x}{2}}+6 \]

Solution is \(y=y_{h}+y_{p}\). The roots of the characteristic equation are \(\pm \frac {1}{2}\),. hence \(y_{h}\) is

\[ y_{h}=c_{1}e^{\frac {1}{2}x}+c_{2}e^{-\frac {1}{2}x}\]

The basis for \(y_{h}\) are \(y_{1}=e^{\frac {1}{2}x},y_{2}=e^{-\frac {1}{2}x}\). Let

\[ y_{p}=y_{1}u_{1}+y_{2}u_{2}\]

Where

\begin{align*} u_{1} & =-\int \frac {y_{2}f\left ( x\right ) }{aW}dx\\ u_{2} & =\int \frac {y_{1}f\left ( x\right ) }{aW}dx \end{align*}

Where \(a=4,f\left ( x\right ) =e^{\frac {x}{2}}+6\) and

\[ W=\begin {vmatrix} y_{1} & y_{2}\\ y_{1}^{\prime } & y_{2}^{\prime }\end {vmatrix} =\begin {vmatrix} e^{\frac {1}{2}x} & e^{-\frac {1}{2}x}\\ \frac {1}{2}e^{\frac {1}{2}x} & -\frac {1}{2}e^{-\frac {1}{2}x}\end {vmatrix} =-\frac {1}{2}-\frac {1}{2}=-1 \]

Hence

\begin{align*} u_{1} & =-\int \frac {e^{-\frac {1}{2}x}\left ( e^{\frac {x}{2}}+6\right ) }{-4}dx=\frac {1}{4}x-3e^{-\frac {1}{2}x}\\ u_{2} & =\int \frac {e^{\frac {1}{2}x}\left ( e^{\frac {x}{2}}+6\right ) }{-4}dx=-\frac {1}{4}e^{\frac {1}{2}x}\left ( e^{\frac {1}{2}x}+12\right ) \end{align*}

Hence

\begin{align*} y_{p} & =y_{1}u_{1}+y_{2}u_{2}\\ & =e^{\frac {1}{2}x}\left ( \frac {1}{4}x-3e^{-\frac {1}{2}x}\right ) +e^{-\frac {1}{2}x}\left ( -\frac {1}{4}e^{\frac {1}{2}x}\left ( e^{\frac {1}{2}x}+12\right ) \right ) \\ & =\frac {1}{4}xe^{\frac {1}{2}x}-\frac {1}{4}e^{\frac {1}{2}x}-6 \end{align*}

Therefore

\begin{align*} y & =y_{h}+y_{p}\\ & =c_{1}e^{\frac {1}{2}x}+c_{2}e^{-\frac {1}{2}x}+\frac {1}{4}xe^{\frac {1}{2}x}-\frac {1}{4}e^{\frac {1}{2}x}-6 \end{align*}

Or by combining terms into new constant, the above becomes

\[ y=c_{3}e^{\frac {1}{2}x}+c_{2}e^{-\frac {1}{2}x}+\frac {1}{4}xe^{\frac {1}{2}x}-6 \]