Q19E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Express the solution to the initial value problem $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{=}\frac{\mathbf{1}}{\mathbf{t}}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{2}$, using definite integrals. Using numerical integration (Appendix C) to approximate the integrals, find an approximation for y(2) to two decimal places.

Step-by-Step Solution

Answer

The general solution is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{1-\mathrm{t}}-{\mathrm{e}}^{\mathrm{t}-1}+\left(\int \frac{{\mathrm{e}}^{-\mathrm{t}}}{2\mathrm{t}}\mathrm{dt}\right){\mathrm{e}}^{\mathrm{t}}+\left(\int -\frac{{\mathrm{e}}^{-\mathrm{t}}}{2\mathrm{t}}\mathrm{dt}\right){\mathrm{e}}^{-\mathrm{t}}$ and the approximation value for y(2)=-1.93.

 

Step 1: Find the particular solution

The differential equation is $\mathrm{y}\text{'}\text{'}-\mathrm{y}=\frac{1}{\mathrm{t}}$.

 

The homogenous equation is ${\mathrm{r}}^2-1=0$.

  

Two independent solutions are $\mathrm{r}=\pm 1$.

 

Then ${\mathrm{y}}_1={\mathrm{e}}^{\mathrm{t}},{\mathrm{y}}_2={\mathrm{te}}^{-\mathrm{t}}$

${\mathrm{y}}_{\mathrm{h}}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{-\mathrm{t}}$ 

The particular solution is ${\mathrm{y}}_{\mathrm{p}}={\mathrm{v}}_1\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}+{\mathrm{v}}_2\left(\mathrm{t}\right)\mathrm{t}{\mathrm{e}}^{-\mathrm{t}}$.

Step 2: Evaluate v 1   and   v 2

Here ${\mathrm{y}}_{\mathrm{p}}={\mathrm{v}}_1\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}+{\mathrm{v}}_2\left(\mathrm{t}\right)\mathrm{t}{\mathrm{e}}^{-\mathrm{t}}$

 

And referring to (9) and solve the system then

$$\begin{gathered} {\mathrm{v}}_1\text{'}{\mathrm{e}}^{\mathrm{t}}+{\mathrm{v}}_2\text{'}{\mathrm{e}}^{-\mathrm{t}}=0 \\ {\mathrm{v}}_1\text{'}{\mathrm{e}}^{\mathrm{t}}-\mathrm{v}{\text{'}}_2{\mathrm{e}}^{-\mathrm{t}}=\frac{\mathrm{f}}{\mathrm{a}} \\ {\mathrm{v}}_1\text{'}{\mathrm{e}}^{\mathrm{t}}-\mathrm{v}{\text{'}}_2{\mathrm{e}}^{-\mathrm{t}}=\frac{1}{\mathrm{t}} \end{gathered}$$

Step 3: Find v 1 ' and v 1

$\begin{array}{rcl}{\mathrm{v}}_1\text{'}& =& \frac{-\mathrm{f}\left(\mathrm{t}\right){\mathrm{y}}_2\left(\mathrm{t}\right)}{\mathrm{a}\left[{\mathrm{y}}_1\left(\mathrm{t}\right)\mathrm{y}{\text{'}}_2\left(\mathrm{t}\right)-\mathrm{y}{\text{'}}_1\left(\mathrm{t}\right){\mathrm{y}}_2\left(\mathrm{t}\right)\right]}\\ & =& \frac{-(\frac{1}{\mathrm{t}}).{\mathrm{e}}^{-\mathrm{t}}}{\left[{\mathrm{e}}^{\mathrm{t}}.(-{\mathrm{e}}^{-\mathrm{t}})-{\mathrm{e}}^{\mathrm{t}}.{\mathrm{e}}^{-\mathrm{t}}\right]}\\ & =& \frac{-(\frac{1}{\mathrm{t}}).{\mathrm{e}}^{-\mathrm{t}}}{-2}\\ & =& \frac{{\mathrm{e}}^{-\mathrm{t}}}{2\mathrm{t}}\end{array}$

Now integrating this.

${\mathrm{v}}_1\left(\mathrm{t}\right)=\int \frac{{\mathrm{e}}^{-\mathrm{t}}}{2\mathrm{t}}\mathrm{dt}$

Step 4: Determine v 2 ' and v 2

$\begin{array}{rcl}{\mathrm{v}}_2\text{'}& =& \frac{\mathrm{f}\left(\mathrm{t}\right){\mathrm{y}}_1\left(\mathrm{t}\right)}{\mathrm{a}\left[{\mathrm{y}}_1\left(\mathrm{t}\right)\mathrm{y}{\text{'}}_2\left(\mathrm{t}\right)-\mathrm{y}{\text{'}}_1\left(\mathrm{t}\right){\mathrm{y}}_2\left(\mathrm{t}\right)\right]}\\ & =& \frac{\left(\frac{1}{\mathrm{t}}\right).{\mathrm{e}}^{\mathrm{t}}}{\left[{\mathrm{e}}^{\mathrm{t}}.(-{\mathrm{e}}^{-\mathrm{t}})-{\mathrm{e}}^{\mathrm{t}}.{\mathrm{e}}^{-\mathrm{t}}\right]}\\ & =& \frac{\left(\frac{1}{\mathrm{t}}\right).{\mathrm{e}}^{-\mathrm{t}}}{-2}\\ & =& -\frac{{\mathrm{e}}^{-\mathrm{t}}}{2\mathrm{t}}\end{array}$

Integrate this.

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

 

Thus, a particular solution is:

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

 

And the general solution is: 

$$\begin{gathered} \mathrm{y}\left(\mathrm{t}\right)={\mathrm{y}}_{\mathrm{h}}\left(\mathrm{t}\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right) \\ \mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{-\mathrm{t}}+\left(\int \frac{{\mathrm{e}}^{-\mathrm{t}}}{2\mathrm{t}}\mathrm{dt}\right){\mathrm{e}}^{\mathrm{t}}+\left(\int -\frac{{\mathrm{e}}^{-\mathrm{t}}}{2\mathrm{t}}\mathrm{dt}\right){\mathrm{e}}^{-\mathrm{t}} \end{gathered}$$

Step 5: Apply initial conditions

The given initial conditions are.

$\begin{array}{rcl}& & \mathrm{y}\left(1\right)={\mathrm{c}}_1{\mathrm{e}}^1+{\mathrm{c}}_2{\mathrm{e}}^{-1}+\left(\int \frac{{\mathrm{e}}^{-1}}{2}\mathrm{dt}\right){\mathrm{e}}^1+\left(\int -\frac{{\mathrm{e}}^{-1}}{2}\mathrm{dt}\right){\mathrm{e}}^{-1}\\ 0& =& {\mathrm{c}}_1\mathrm{e}+{\mathrm{c}}_2{\mathrm{e}}^{-1}\\ {\mathrm{c}}_2& =& -{\mathrm{c}}_1{\mathrm{e}}^2\end{array}$

  

And 

$$\begin{gathered} \mathrm{y}\text{'}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{t}}-{\mathrm{c}}_2{\mathrm{e}}^{-\mathrm{t}}+\left(\int \frac{{\mathrm{e}}^{-\mathrm{t}}}{2\mathrm{t}}\mathrm{dt}\right){\mathrm{e}}^{\mathrm{t}}+\frac{1}{2\mathrm{t}}+\left(\int \frac{{\mathrm{e}}^{-\mathrm{t}}}{2\mathrm{t}}\mathrm{dt}\right){\mathrm{e}}^{-\mathrm{t}}-\frac{1}{2\mathrm{t}} \\ \mathrm{y}\text{'}\left(1\right)={\mathrm{c}}_1\mathrm{e}-{\mathrm{c}}_2{\mathrm{e}}^{-1}+\left(\int \frac{{\mathrm{e}}^{-1}}{2}\mathrm{dt}\right)\mathrm{e}+\frac{1}{2}+\left(\int \frac{{\mathrm{e}}^{-1}}{2}\mathrm{dt}\right){\mathrm{e}}^{-1}-\frac{1}{2} \\ -2={\mathrm{c}}_1\mathrm{e}-{\mathrm{c}}_2{\mathrm{e}}^{-1} \end{gathered}$$

 

Solving for ${\mathrm{c}}_1 \mathrm{and} {\mathrm{c}}_2$.

${\mathrm{c}}_1=-\frac{1}{\mathrm{e}} \mathrm{and} {\mathrm{c}}_2=\mathrm{e}$

There for the solution is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{1-\mathrm{t}}-{\mathrm{e}}^{\mathrm{t}-1}+\left(\int \frac{{\mathrm{e}}^{-\mathrm{t}}}{2\mathrm{t}}\mathrm{dt}\right){\mathrm{e}}^{\mathrm{t}}+\left(\int -\frac{{\mathrm{e}}^{-\mathrm{t}}}{2\mathrm{t}}\mathrm{dt}\right){\mathrm{e}}^{-\mathrm{t}}$.

And

$\begin{array}{rcl}& & \mathrm{y}\left(2\right)={\mathrm{e}}^{1-2}-{\mathrm{e}}^{2-1}+\left(\int \frac{{\mathrm{e}}^{-2}}{2\left(2\right)}\mathrm{dt}\right){\mathrm{e}}^2+\left(\int -\frac{{\mathrm{e}}^{-2}}{2\left(2\right)}\mathrm{dt}\right){\mathrm{e}}^{-2}\\ & =& {\mathrm{e}}^{-1}-\mathrm{e}\\ & =& -1.93\end{array}$

 

This is the required result.