91Ó°ÊÓ

Now going to take a brief detour and look at solutions to non-constant coefficient, second order differential equations of the form.

$\mathbf{p}\left(\mathbf{t}\right)\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{q}\left(\mathbf{t}\right)\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{r}\left(\mathbf{t}\right)\mathbf{y}\mathbf{=}\mathbf{0}$

In general, finding solutions to these kinds of differential equations can be much more difficult than finding solutions to constant coefficient differential equations. This method is called reduction of order.

Given function $\mathbf{f}\left(\mathbf{x}\right)\mathbf{=}{\mathbf{e}}^{2x}$,is a solution to

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{0}\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}\mathrm{......}\left(\mathbf{1}\right)$

And

$\mathbf{y}\left(\mathbf{x}\right)\mathbf{=}\mathbf{v}\left(\mathbf{x}\right)\mathbf{f}\left(\mathbf{x}\right)\mathbf{=}\mathbf{v}\left(\mathbf{x}\right){\mathbf{e}}^{2x}$

Now find the derivative of y for equation (1),

$\begin{array}{l}\mathbf{y}\mathbf{=}{\mathbf{ve}}^{2x}\\ \mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{2}{\mathbf{ve}}^{2x}\mathbf{+}\mathbf{v}\mathbf{\text{'}}{\mathbf{e}}^{2x}\\ \mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\mathbf{4}{\mathbf{ve}}^{2x}\mathbf{+}\mathbf{2}\mathbf{v}\mathbf{\text{'}}{\mathbf{e}}^{2x}\mathbf{+}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}{\mathbf{e}}^{2x}\mathbf{+}\mathbf{2}\mathbf{v}\mathbf{\text{'}}{\mathbf{e}}^{2x}\\ \mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\mathbf{4}{\mathbf{ve}}^{2x}\mathbf{+}\mathbf{4}\mathbf{v}\mathbf{\text{'}}{\mathbf{e}}^{2x}\mathbf{+}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}{\mathbf{e}}^{2x}\\ \mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\mathbf{8}{\mathbf{ve}}^{2x}\mathbf{+}\mathbf{4}\mathbf{v}\mathbf{\text{'}}{\mathbf{e}}^{2x}\mathbf{+}\mathbf{4}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}{\mathbf{e}}^{2x}\mathbf{+}\mathbf{8}\mathbf{v}\mathbf{\text{'}}{\mathbf{e}}^{2x}\mathbf{+}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}{\mathbf{e}}^{2x}\mathbf{+}\mathbf{2}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}{\mathbf{e}}^{2x}\\ \mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\mathbf{8}{\mathbf{ve}}^{2x}\mathbf{+}\mathbf{12}\mathbf{v}\mathbf{\text{'}}{\mathbf{e}}^{2x}\mathbf{+}\mathbf{6}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}{\mathbf{e}}^{2x}\mathbf{+}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}{\mathbf{e}}^{2x}\end{array}$

Substitute all values in the equation (1),

$\begin{array}{c}y\text{'}\text{'}\text{'}-2y\text{'}\text{'}+y\text{'}-2y=0\\ \mathbf{8}{\mathbf{ve}}^{2x}\mathbf{+}\mathbf{12}\mathbf{v}\mathbf{\text{'}}{\mathbf{e}}^{2x}\mathbf{+}\mathbf{6}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}{\mathbf{e}}^{2x}\mathbf{+}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}{\mathbf{e}}^{2x}\mathbf{-}\mathbf{2}(\mathbf{4}{\mathbf{ve}}^{2x}\mathbf{+}\mathbf{4}\mathbf{v}\mathbf{\text{'}}{\mathbf{e}}^{2x}\mathbf{+}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}{\mathbf{e}}^{2x})\mathbf{+}\mathbf{2}{\mathbf{ve}}^{2x}\mathbf{+}\mathbf{v}\mathbf{\text{'}}{\mathbf{e}}^{2x}\mathbf{-}\mathbf{2}{\mathbf{ve}}^{2x}\mathbf{=}\mathbf{0}\\ \mathbf{8}{\mathbf{ve}}^{2x}\mathbf{+}\mathbf{12}\mathbf{v}\mathbf{\text{'}}{\mathbf{e}}^{2x}\mathbf{+}\mathbf{6}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}{\mathbf{e}}^{2x}\mathbf{+}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}{\mathbf{e}}^{2x}\mathbf{-}\mathbf{8}{\mathbf{ve}}^{2x}\mathbf{-}\mathbf{8}\mathbf{v}\mathbf{\text{'}}{\mathbf{e}}^{2x}\mathbf{-}\mathbf{2}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}{\mathbf{e}}^{2x}\mathbf{+}\mathbf{v}\mathbf{\text{'}}{\mathbf{e}}^{2x}\mathbf{=}\mathbf{0}\\ \mathbf{5}\mathbf{v}\mathbf{\text{'}}{\mathbf{e}}^{2x}\mathbf{+}\mathbf{4}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}{\mathbf{e}}^{2x}\mathbf{+}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}{\mathbf{e}}^{2x}\mathbf{=}\mathbf{0}\\ (\mathbf{5}\mathbf{v}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}){\mathbf{e}}^{2x}\mathbf{=}\mathbf{0}\\ 5v\text{'}+4v\text{'}\text{'}+v\text{'}\text{'}\text{'}=0\end{array}$

Use the value$\mathbf{w}\mathbf{=}\mathbf{v}\mathbf{\text{'}}$ in the above expression,

$\begin{array}{l}5w+4w\text{'}+w\text{'}\text{'}=0\\ w\text{'}\text{'}+4w\text{'}+5w=0\end{array}$

Solve the above equation for w,

$\begin{array}{c}({\mathbf{D}}^2\mathbf{+}\mathbf{4}\mathbf{D}\mathbf{+}\mathbf{5})\mathbf{w}\mathbf{=}\mathbf{0}\\ {\mathbf{D}}^2\mathbf{+}\mathbf{4}\mathbf{D}\mathbf{+}\mathbf{5}=0\\ \mathbf{D}\mathbf{=}\frac{\mathbf{-}\mathbf{4}\mathbf{Â\pm }\sqrt{16-20}}{2}\\ \mathbf{D}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{Â\pm }\frac{\sqrt{-4}}{2}\end{array}$

The solution of w is,

$\begin{array}{c}\mathbf{w}\left(\mathbf{x}\right)\mathbf{=}{\mathbf{Ae}}^{-2x}\mathbf{cosx}\mathbf{+}{\mathbf{Be}}^{-2x}\mathbf{sinx}\\ \mathbf{v}\mathbf{\text{'}}\mathbf{=}{\mathbf{Ae}}^{-2x}\mathbf{cosx}\mathbf{+}{\mathbf{Be}}^{-2x}\mathbf{sinx}\end{array}$

Integrating both sides with respect to x,

$\begin{array}{c}∫\mathbf{v}\mathbf{\text{'}}\mathbf{=}∫({\mathbf{Ae}}^{-2x}\mathbf{cosx}\mathbf{+}{\mathbf{Be}}^{-2x}\mathbf{sinx})\mathbf{dx}\\ \mathbf{v}\mathbf{=}\mathbf{A}\frac{{\mathbf{e}}^{-2x}}{5}(\mathbf{-}\mathbf{2}\mathbf{cosx}\mathbf{+}\mathbf{sinx})\mathbf{+}\mathbf{B}\frac{{\mathbf{e}}^{-2x}}{5}(\mathbf{-}\mathbf{2}\mathbf{sinx}\mathbf{-}\mathbf{cosx})\\ \mathbf{v}\mathbf{=}{\mathbf{v}}_1\mathbf{+}{\mathbf{v}}_2\end{array}$

We have,

$\begin{array}{c}{\mathbf{y}}_i\mathbf{=}{\mathbf{v}}_i\mathbf{f}\mathbf{=}\frac{{\mathbf{v}}_i}{{\mathbf{e}}^{-2x}}\\ {\mathbf{y}}_1\mathbf{=}\mathbf{A}\frac{{\mathbf{e}}^{-2x}}{\mathbf{5}{\mathbf{e}}^{-2x}}(\mathbf{-}\mathbf{2}\mathbf{cosx}\mathbf{+}\mathbf{sinx})\mathbf{=}\frac{A}{5}(\mathbf{-}\mathbf{2}\mathbf{cosx}\mathbf{+}\mathbf{sinx})\\ {\mathbf{y}}_2\mathbf{=}\mathbf{B}\frac{{\mathbf{e}}^{-2x}}{\mathbf{5}{\mathbf{e}}^{-2x}}(\mathbf{-}\mathbf{2}\mathbf{sinx}\mathbf{-}\mathbf{cosx})\mathbf{=}\frac{B}{5}(\mathbf{-}\mathbf{2}\mathbf{sinx}\mathbf{-}\mathbf{cosx})\\ {\mathbf{y}}_3\mathbf{=}{\mathbf{e}}^{2x}\end{array}$

Hence, ${\mathbf{y}}_1\mathbf{,}\text{ }{\mathbf{y}}_2$and${\mathbf{y}}_3$ are linearly independent solutions.

${\mathbf{y}}_1\mathbf{=}\frac{(\mathbf{sinx}\mathbf{-}\mathbf{2}\mathbf{cosx})}{5},\text{ â¶Ä‰}{\mathbf{y}}_2\mathbf{=}\frac{(\mathbf{-}\mathbf{2}\mathbf{sinx}\mathbf{-}\mathbf{cosx})}{5},\text{ â¶Ä‰}\mathbf{f}\mathbf{=}{\mathbf{e}}^{2x}$