Q21E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Suppose y satisfies the equation $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{10}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{25}\mathbf{y}\mathbf{=}{\mathbf{e}}^{{\mathbf{t}}^{\mathbf{3}}}$ subject to $\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{andy}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{5}$ Estimate y(0.2) to within $\mathbf{\pm }\mathbf{0}\mathbf{.}\mathbf{0001}$ by numerically approximating the integrals in the variation of parameters formula.

Step-by-Step Solution

Answer

The general solution is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{-5\mathrm{t}}+10{\mathrm{e}}^{-5\mathrm{t}}-{\mathrm{e}}^{-5\mathrm{t}}\int {\mathrm{te}}^{{\mathrm{t}}^3+5\mathrm{t}}\mathrm{dt}+{\mathrm{te}}^{-5\mathrm{t}}\int {\mathrm{te}}^{{\mathrm{t}}^3+5\mathrm{t}}\mathrm{dt}$.

Step 1: Find a particular solution

The homogenous equation is ${\mathrm{r}}^2+10\mathrm{r}+25=0$.

 

Two independent solutions are $\mathrm{r}=-5,-5$.

 

Then ${\mathrm{y}}_1={\mathrm{e}}^{-5\mathrm{t}},{\mathrm{y}}_2={\mathrm{te}}^{-5\mathrm{t}}$

${\mathrm{y}}_{\mathrm{h}}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{-5\mathrm{t}}+{\mathrm{c}}_2{\mathrm{te}}^{-5\mathrm{t}}$

The particular solution is ${\mathrm{y}}_{\mathrm{p}}={\mathrm{v}}_1\left(\mathrm{t}\right){\mathrm{e}}^{-5\mathrm{t}}+{\mathrm{v}}_2\left(\mathrm{t}\right)\mathrm{t}{\mathrm{e}}^{-5\mathrm{t}}$ 

Step 2: Evaluate v 1     and     v 2

Here ${\mathrm{y}}_{\mathrm{p}}={\mathrm{v}}_1\left(\mathrm{t}\right){\mathrm{e}}^{-5\mathrm{t}}+{\mathrm{v}}_2\left(\mathrm{t}\right)\mathrm{t}{\mathrm{e}}^{-5\mathrm{t}}$

 

And referring to (9) and solve the system then

$\begin{array}{rcl}& & -5{\mathrm{v}}_1\text{'}{\mathrm{e}}^{-5\mathrm{t}}+{\mathrm{v}}_2\text{'}(-5\mathrm{t}{\mathrm{e}}^{-5\mathrm{t}}+{\mathrm{e}}^{-5\mathrm{t}})=0\\ & & -5{\mathrm{v}}_1\text{'}{\mathrm{e}}^{-5\mathrm{t}}-\mathrm{v}{\text{'}}_2(-5\mathrm{t}{\mathrm{e}}^{-5\mathrm{t}}+{\mathrm{e}}^{-5\mathrm{t}})=\frac{\mathrm{f}}{\mathrm{a}}\\ & & -5{\mathrm{v}}_1\text{'}{\mathrm{e}}^{-5\mathrm{t}}-\mathrm{v}{\text{'}}_2(-5\mathrm{t}{\mathrm{e}}^{-5\mathrm{t}}+{\mathrm{e}}^{-5\mathrm{t}})={\mathrm{e}}^{{\mathrm{t}}^3}\end{array}$ 

Step 3: Find v 1 ' and v 1

$\begin{array}{rcl}{\mathrm{v}}_1\text{'}& =& \frac{-\mathrm{f}\left(\mathrm{t}\right){\mathrm{y}}_2\left(\mathrm{t}\right)}{\mathrm{a}\left[{\mathrm{y}}_1\left(\mathrm{t}\right)\mathrm{y}{\text{'}}_2\left(\mathrm{t}\right)-\mathrm{y}{\text{'}}_1\left(\mathrm{t}\right){\mathrm{y}}_2\left(\mathrm{t}\right)\right]}\\ & =& \frac{-{\mathrm{e}}^{{\mathrm{t}}^3}.{\mathrm{te}}^{-5\mathrm{t}}}{\left[{\mathrm{e}}^{-5\mathrm{t}}.(-5\mathrm{t}{\mathrm{e}}^{-5\mathrm{t}}+{\mathrm{e}}^{-5\mathrm{t}})-(-5{\mathrm{e}}^{-5\mathrm{t}}.{\mathrm{te}}^{-5\mathrm{t}})\right]}\\ & =& \frac{-{\mathrm{e}}^{{\mathrm{t}}^3}.{\mathrm{te}}^{-5\mathrm{t}}}{{\mathrm{e}}^{-10\mathrm{t}}}\\ & =& -{\mathrm{te}}^{{\mathrm{t}}^3+5\mathrm{t}}\end{array}$ 

Now integrating this.

${\mathrm{v}}_1\left(\mathrm{t}\right)=\int -{\mathrm{te}}^{{\mathrm{t}}^3+5\mathrm{t}}\mathrm{dt}$

Step 4: Determine v 2 ' and v 2

$\begin{array}{rcl}{\mathrm{v}}_2\text{'}& =& \frac{\mathrm{f}\left(\mathrm{t}\right){\mathrm{y}}_1\left(\mathrm{t}\right)}{\mathrm{a}\left[{\mathrm{y}}_1\left(\mathrm{t}\right)\mathrm{y}{\text{'}}_2\left(\mathrm{t}\right)-\mathrm{y}{\text{'}}_1\left(\mathrm{t}\right){\mathrm{y}}_2\left(\mathrm{t}\right)\right]}\\ & =& \frac{-{\mathrm{e}}^{{\mathrm{t}}^3}.{\mathrm{e}}^{-5\mathrm{t}}}{\left[{\mathrm{e}}^{-5\mathrm{t}}.(-5\mathrm{t}{\mathrm{e}}^{-5\mathrm{t}}+{\mathrm{e}}^{-5\mathrm{t}})-(-5{\mathrm{e}}^{-5\mathrm{t}}.{\mathrm{te}}^{-5\mathrm{t}})\right]}\\ & =& \frac{-{\mathrm{e}}^{{\mathrm{t}}^3}.{\mathrm{te}}^{-5\mathrm{t}}}{{\mathrm{e}}^{-10\mathrm{t}}}\\ & =& {\mathrm{te}}^{{\mathrm{t}}^3+5\mathrm{t}}\end{array}$

 

Integrate this.

${\mathrm{v}}_2\left(\mathrm{t}\right)=\int {\mathrm{te}}^{{\mathrm{t}}^3+5\mathrm{t}}\mathrm{dt}$

Thus, a particular solution is:

 ${\mathrm{y}}_{\mathrm{p}}=-{\mathrm{e}}^{-5\mathrm{t}}\int {\mathrm{te}}^{{\mathrm{t}}^3+5\mathrm{t}}\mathrm{dt}+{\mathrm{te}}^{-5\mathrm{t}}\int {\mathrm{te}}^{{\mathrm{t}}^3+5\mathrm{t}}\mathrm{dt}$

And the general solution is: 

$$\begin{gathered} \mathrm{y}\left(\mathrm{t}\right)={\mathrm{y}}_{\mathrm{h}}\left(\mathrm{t}\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right) \\ \mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{-5\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{-5\mathrm{t}}-{\mathrm{e}}^{-5\mathrm{t}}\int {\mathrm{te}}^{{\mathrm{t}}^3+5\mathrm{t}}\mathrm{dt}+{\mathrm{te}}^{-5\mathrm{t}}\int {\mathrm{te}}^{{\mathrm{t}}^3+5\mathrm{t}}\mathrm{dt} \end{gathered}$$ 

Step 5: Apply initial conditions.

The given initial conditions are $\mathrm{y}\left(0\right)=1 \mathrm{and} \mathrm{y}\text{'}\left(0\right)=-5$.

$$\begin{gathered} \mathrm{y}\left(0\right)={\mathrm{c}}_1{\mathrm{e}}^0+{\mathrm{c}}_2{\mathrm{e}}^0-{\mathrm{e}}^0\int 0.{\mathrm{e}}^{0^3+0}\mathrm{dt}+0.{\mathrm{e}}^0\int 0{\mathrm{e}}^{0^3+0}\mathrm{dt} \\ 1={\mathrm{c}}_1 \end{gathered}$$

 

And

$$\begin{gathered} \mathrm{y}\text{'}\left(\mathrm{t}\right)=-5{\mathrm{c}}_1{\mathrm{e}}^{-5\mathrm{t}}+{\mathrm{c}}_2(-5\mathrm{t}{\mathrm{e}}^{-5\mathrm{t}}+{\mathrm{e}}^{-5\mathrm{t}})+5{\mathrm{e}}^{-5\mathrm{t}}\int {\mathrm{te}}^{{\mathrm{t}}^3+5\mathrm{t}}\mathrm{dt}+{\mathrm{e}}^{-5\mathrm{t}}(\mathrm{t}{\mathrm{e}}^{{\mathrm{t}}^3+5\mathrm{t}})+{\mathrm{e}}^{-5\mathrm{t}}\int {\mathrm{te}}^{{\mathrm{t}}^3+5\mathrm{t}}\mathrm{dt}-5{\mathrm{te}}^{-5\mathrm{t}}\int {\mathrm{te}}^{{\mathrm{t}}^3+5\mathrm{t}}\mathrm{dt} \\ +{\mathrm{te}}^{-5\mathrm{t}}\int {\mathrm{te}}^{{\mathrm{t}}^3+5\mathrm{t}}\mathrm{dt} \\ -5=-5{\mathrm{c}}_1+{\mathrm{c}}_2 \\ -5=-5\left(1\right)+{\mathrm{c}}_2 \\ {\mathrm{c}}_2=10 \end{gathered}$$

Therefore, the solution is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{-5\mathrm{t}}+10{\mathrm{e}}^{-5\mathrm{t}}-{\mathrm{e}}^{-5\mathrm{t}}\int {\mathrm{te}}^{{\mathrm{t}}^3+5\mathrm{t}}\mathrm{dt}+{\mathrm{te}}^{-5\mathrm{t}}\int {\mathrm{te}}^{{\mathrm{t}}^3+5\mathrm{t}}\mathrm{dt}$.

 

This is the required result.