Solve the integral equation $y(t)={e}^{t}(1+{\int}_{0}^{t}{e}^{-\tau}y(\tau )d\tau )$ with Laplace transform

Jacoby Erickson
2022-10-13
Answered

Solve the integral equation $y(t)={e}^{t}(1+{\int}_{0}^{t}{e}^{-\tau}y(\tau )d\tau )$ with Laplace transform

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Messiah Trevino

Answered 2022-10-14
Author has **18** answers

Let the Laplace of y(t) be Y(s).

Applying Laplace on both the sides we end up with

$$\begin{array}{rl}>Y(s)& =\frac{1}{s-1}+(L[{e}^{t}])(L[y(t)])\end{array}$$

$$\begin{array}{rl}>Y(s)& =\frac{1}{s-1}+\frac{1}{s-1}Y(s)\end{array}$$

Solving we end up in,

$$\begin{array}{rl}>Y(s)& =\frac{1}{s-2}\end{array}$$

Taking inverse Laplace we get, $$y(t)={e}^{2t}$$

Applying Laplace on both the sides we end up with

$$\begin{array}{rl}>Y(s)& =\frac{1}{s-1}+(L[{e}^{t}])(L[y(t)])\end{array}$$

$$\begin{array}{rl}>Y(s)& =\frac{1}{s-1}+\frac{1}{s-1}Y(s)\end{array}$$

Solving we end up in,

$$\begin{array}{rl}>Y(s)& =\frac{1}{s-2}\end{array}$$

Taking inverse Laplace we get, $$y(t)={e}^{2t}$$

raapjeqp

Answered 2022-10-15
Author has **2** answers

$$y(t)={e}^{t}{\textstyle (}1+{\int}_{0}^{t}{e}^{-\tau}\text{}y(\tau )d\tau {\textstyle )}$$

$${y}^{\prime}(t)={e}^{t}{\textstyle (}1+{\int}_{0}^{t}{e}^{-\tau}\text{}y(\tau )d\tau {\textstyle )}+y(t)=2y(t)\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}y(t)=C{e}^{2t}$$

Substitute into the original equation to find $C=1$

$$C{e}^{2t}={e}^{t}{\textstyle (}1+C{\int}_{0}^{t}{e}^{\tau}d\tau {\textstyle )}$$

$$C{e}^{t}=1+C({e}^{t}-1)\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}C=1$$

$${y}^{\prime}(t)={e}^{t}{\textstyle (}1+{\int}_{0}^{t}{e}^{-\tau}\text{}y(\tau )d\tau {\textstyle )}+y(t)=2y(t)\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}y(t)=C{e}^{2t}$$

Substitute into the original equation to find $C=1$

$$C{e}^{2t}={e}^{t}{\textstyle (}1+C{\int}_{0}^{t}{e}^{\tau}d\tau {\textstyle )}$$

$$C{e}^{t}=1+C({e}^{t}-1)\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}C=1$$

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$y=6{c}_{1}-2{c}_{2}\mathrm{exp}(-4t)+{3}_{{c}_{3}}\mathrm{exp}\left(3t\right)$

where D is the differential operator, how to get the general solution of y? The solution suggest that it is

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I worked out this far and I am having trouble with the integral

$$F(s)={\int}_{0}^{\mathrm{\infty}}{e}^{-st}f(t)dt$$

$$f(t)={e}^{t+7}$$

$${\int}_{0}^{\mathrm{\infty}}{e}^{-(s-1)t}{e}^{7}dt$$

we can say that ${e}^{7}={c}_{1}$ and show

$$-\frac{{c}_{1}}{s-1}{\int}_{0}^{\mathrm{\infty}}{e}^{-(s-1)t}dt$$

I would surmise that I can take the limit at this point. So

$$-\frac{{c}_{1}}{s-1}\underset{b}{\overset{\mathrm{\infty}}{lim}}{\int}_{0}^{b}{e}^{-(s-1)t}dt$$

which is likely evaluates as

$$-\frac{{c}_{1}}{s-1}(0-1)$$

yielding

$$\frac{{e}^{7}}{s-1}$$

What is "good practice" that I have not shown?

$$F(s)={\int}_{0}^{\mathrm{\infty}}{e}^{-st}f(t)dt$$

$$f(t)={e}^{t+7}$$

$${\int}_{0}^{\mathrm{\infty}}{e}^{-(s-1)t}{e}^{7}dt$$

we can say that ${e}^{7}={c}_{1}$ and show

$$-\frac{{c}_{1}}{s-1}{\int}_{0}^{\mathrm{\infty}}{e}^{-(s-1)t}dt$$

I would surmise that I can take the limit at this point. So

$$-\frac{{c}_{1}}{s-1}\underset{b}{\overset{\mathrm{\infty}}{lim}}{\int}_{0}^{b}{e}^{-(s-1)t}dt$$

which is likely evaluates as

$$-\frac{{c}_{1}}{s-1}(0-1)$$

yielding

$$\frac{{e}^{7}}{s-1}$$

What is "good practice" that I have not shown?

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${y}^{\u2033}-2{y}^{\prime}3y=3{e}^{t};y(0)=-1,{y}^{\prime}(0)=2$

${y}^{\u2033}-2{y}^{\prime}3y=3{e}^{t};y(0)=-1,{y}^{\prime}(0)=2$

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My try:

We get Laplace transform of

$g(t)={t}^{a}$

is:

$\hat{g}(t)=1/{s}^{a+1}{\int}_{0}^{\mathrm{\infty}}{e}^{-t}{x}^{a}$

then I was stuck at question 2) which asks me to evaluate the inverse laplace transform of $\hat{g}(p)$ which is

$1/2\pi i{\int}_{0}^{\mathrm{\infty}}{e}^{pt}\hat{g}(p)dp$

I know the answer should be ${t}^{a}$ as the inverse transform comes back to itself, but I cannot figure out how to make the contour integration. I tried to apply Cauchy's residue theorem to eliminate the $1/2\pi i$ but was stuck then.

My try:

We get Laplace transform of

$g(t)={t}^{a}$

is:

$\hat{g}(t)=1/{s}^{a+1}{\int}_{0}^{\mathrm{\infty}}{e}^{-t}{x}^{a}$

then I was stuck at question 2) which asks me to evaluate the inverse laplace transform of $\hat{g}(p)$ which is

$1/2\pi i{\int}_{0}^{\mathrm{\infty}}{e}^{pt}\hat{g}(p)dp$

I know the answer should be ${t}^{a}$ as the inverse transform comes back to itself, but I cannot figure out how to make the contour integration. I tried to apply Cauchy's residue theorem to eliminate the $1/2\pi i$ but was stuck then.

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