Q44E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a second linearly independent solution using reduction of order.

$\mathbf{ty}\text{'}\text{'}\mathbf{+}\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{2}\mathbf{t}\mathbf{)}\mathbf{y}\text{'}\mathbf{+}\mathbf{(}\mathbf{t}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{,} \mathbf{t}\mathbf{>}\mathbf{0}\mathbf{;} \mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\mathbf{t}}$

Step-by-Step Solution

Answer

The second linearly independent solution of the given equation:

$\mathrm{ty}\text{'}\text{'}+(1-2\mathrm{t})\mathrm{y}\text{'}+(\mathrm{t}-1)\mathrm{y}=0, \mathrm{t}>0; \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{t}}$ is $\mathrm{y}={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{\mathrm{t}}\mathrm{lnt}$.

Step 1: Finding y

Given differential equation is $\mathrm{ty}\text{'}\text{'}+(1-2\mathrm{t})\mathrm{y}\text{'}+(\mathrm{t}-1)\mathrm{y}=0$ then the standard form of the solution is $\mathrm{y}\text{'}\text{'}+\frac{1-2\mathrm{t}}{\mathrm{t}}\mathrm{y}\text{'}+\frac{\mathrm{t}-1}{\mathrm{t}}\mathrm{x}=0$ and $\mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{t}}$ is one of the solutions and $\mathrm{p}\left(\mathrm{t}\right)=\frac{1-2\mathrm{t}}{\mathrm{t}}$.

 

Then:

${\mathrm{y}}_2={\mathrm{y}}_1\left(\mathrm{t}\right)\int \frac{{\mathrm{e}}^{-\int \mathrm{p}\left(\mathrm{t}\right)}\mathrm{dt}}{{\mathrm{y}}_1^2\left(\mathrm{t}\right)}\mathrm{dt}$.

Step 2: Simplification

Simplify the above

$\begin{array}{rcl}d\mathrm{y}& =& {\mathrm{e}}^{\mathrm{t}}\int \frac{{\mathrm{e}}^{-\int \frac{1-2\mathrm{t}}{\mathrm{t}}}\mathrm{dt}}{{\left({\mathrm{e}}^{\mathrm{t}}\right)}^2}\mathrm{dt}\\ & =& {\mathrm{e}}^{\mathrm{t}}\int \frac{{\mathrm{e}}^{2\mathrm{t}-\ln \left(\mathrm{t}\right)}}{{\mathrm{e}}^{2\mathrm{t}}}\mathrm{dt}\\ & =& {\mathrm{e}}^{\mathrm{t}}\times \int \frac{{\mathrm{t}}^{-1}\times {\mathrm{e}}^{2\mathrm{t}}}{{\mathrm{e}}^{2\mathrm{t}}}\mathrm{dt}\\ \mathrm{y}& =& {\mathrm{lnte}}^{\mathrm{t}}\end{array}$ 

So, the solution is $\mathrm{y}={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{\mathrm{t}}\mathrm{lnt}$.