4.3.1.2 Solved by finding roots of characteristic equation

ode internal name "second order linear constant coeff"

These are solved by finding roots of characteristic equation. This is the standard method. Homogeneous and inhomogeneous. The method of Variation of parameters and the method of undetermined coefficients are both used to find the particular solution. If hint "laplace" is given, then the ODE is solved using Laplace transform method. If hint "series" is given then series method is used.

4.3.1.2.1 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

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