Q43E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a second linearly independent solution using reduction of order.

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

Step-by-Step Solution

Answer

The second linearly independent solution of the given equation 

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

Step 1: Finding y 2

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

$\begin{array}{rcl}{\mathrm{y}}_2& =& {\mathrm{y}}_1\left(\mathrm{t}\right)\int \frac{{\mathrm{e}}^{-\int \mathrm{p}\left(\mathrm{t}\right)}\mathrm{dt}}{{\mathrm{y}}_1^2\left(\mathrm{t}\right)}\mathrm{dt}\\ & =& {\mathrm{e}}^{\mathrm{t}}\int \frac{{\mathrm{e}}^{\int \frac{1+\mathrm{t}}{\mathrm{t}}}\mathrm{dt}}{{\left({\mathrm{e}}^{\mathrm{t}}\right)}^2}\mathrm{dt}\\ & =& {\mathrm{e}}^{\mathrm{t}}\int \frac{{\mathrm{e}}^{\ln \left({\mathrm{t}}^1\right)+\mathrm{t}}}{{\mathrm{e}}^{2\mathrm{t}}}\end{array}$

 

Step 2: Simplification

Simplify the above

$$\begin{gathered} ={\mathrm{e}}^{\mathrm{t}}\times \int \frac{\mathrm{t}\times {\mathrm{e}}^{\mathrm{t}}}{{\mathrm{e}}^{2\mathrm{t}}} \\ ={\mathrm{e}}^{\mathrm{t}}\times \frac{(-\mathrm{t}-1){\mathrm{e}}^{-\mathrm{t}}}{{\mathrm{e}}^{2\mathrm{t}}} \\ =-\frac{\mathrm{t}+1}{{\mathrm{e}}^{2\mathrm{t}}} \end{gathered}$$

So, the solution is $\mathrm{y}={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{t}}-{\mathrm{c}}_2\frac{\mathrm{t}+1}{{\mathrm{e}}^{2\mathrm{t}}}$.