# In an integro-differential equation, the unknown dependent variable y appears within an integral, and its derivative frac{dy}{dt} also appears. Consider the following initial value problem, defined for t > 0: frac{{{left.{d}{y}right.}}}{{{left.{d}{t}right.}}}+{4}{int_{{0}}^{{t}}}{y}{left({t}-{w}right)}{e}^{{-{4}{w}}}{d}{w}={3},{y}{left({0}right)}={0} a) Use convolution and Laplace transforms to find the Laplace transform of the solution. {Y}{left({s}right)}={L}{leftlbrace{y}{left({t}right)}right)}{rbrace}-? b) Obtain the solution y(t). y(t) - ?

Question
Laplace transform
In an integro-differential equation, the unknown dependent variable y appears within an integral, and its derivative $$\frac{dy}{dt}$$ also appears. Consider the following initial value problem, defined for t > 0:
$$\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}+{4}{\int_{{0}}^{{t}}}{y}{\left({t}-{w}\right)}{e}^{{-{4}{w}}}{d}{w}={3},{y}{\left({0}\right)}={0}$$
a) Use convolution and Laplace transforms to find the Laplace transform of the solution.
$${Y}{\left({s}\right)}={L}{\left\lbrace{y}{\left({t}\right)}\right)}{\rbrace}-?$$
b) Obtain the solution y(t).
y(t) - ?

2021-02-10
Step 1
Given, $$\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}+{4}{\int_{{0}}^{{t}}}{y}{\left({t}-{w}\right)}{e}^{{-{4}{w}}}{d}{w}={3},{y}{\left({0}\right)}={0}$$
a) Taking Laplace transform of the above equation
$${L}{\left({y}'\right)}+{4}{L}{\left({\int_{{0}}^{{t}}}{y}{\left({t}-{w}\right)}{e}^{{-{4}{w}}}{d}{w}\right)}={L}{\left({3}\right)}$$
$${s}{y}{\left({s}\right)}-{y}{\left({0}\right)}+{4}\frac{{{y}{\left({s}\right)}}}{{{s}+{4}}}=\frac{3}{{s}}$$
$${y}{\left({s}\right)}{\left[{s}+\frac{4}{{{s}+{4}}}\right]}=\frac{3}{{s}}$$
$${y}{\left({s}\right)}{\left[\frac{{{s}^{2}+{4}{s}+{4}}}{{{s}+{4}}}\right]}=\frac{3}{{s}}$$
$${y}{\left({s}\right)}=\frac{3}{{s}}{\left[\frac{{{s}+{4}}}{{{s}^{2}+{4}{s}+{4}}}\right]}$$
$$=\frac{3}{{s}}{\left[\frac{{{s}+{4}}}{{{\left({s}+{2}\right)}^{2}}}\right]}$$
$${y}{\left({s}\right)}=\frac{{{3}{\left({s}+{4}\right)}}}{{{s}{\left({s}+{2}\right)}^{2}}}$$
Step 2
Now take, $$\frac{{{3}{\left({s}+{4}\right)}}}{{{s}{\left({s}+{2}\right)}^{2}}}$$ resolve into partial fractions
$$\frac{{{3}{\left({s}+{4}\right)}}}{{{s}{\left({s}+{2}\right)}^{2}}}=\frac{A}{{s}}+\frac{B}{{{s}+{2}}}+\frac{C}{{\left({s}+{2}\right)}^{2}}$$
$$\frac{{{3}{\left({s}+{4}\right)}}}{{{s}{\left({s}+{2}\right)}^{2}}}=\frac{{{A}{\left({s}+{2}\right)}^{2}+{B}{s}{\left({s}+{2}\right)}+{C}{s}}}{{{s}{\left({s}+{2}\right)}^{2}}}$$
$${3}{\left({s}+{4}\right)}={A}{\left({s}+{2}\right)}^{2}+{B}{s}{\left({s}+{2}\right)}+{C}{s}$$
Put $$s=0$$
$$12=4A+0+0$$
A=3
Put $$s=-2$$
$$6=0+0-2C$$
$$C=-3$$
Consider coefficients of 's2'
$$0=A+B$$
$$A=-B$$ or $$B=-A$$
$$B=-3$$
Put the values in the equation
$$\frac{{{3}{\left({s}+{4}\right)}}}{{{s}{\left({s}+{2}\right)}^{2}}}=\frac{3}{{s}}+\frac{{-{3}}}{{{s}+{2}}}+\frac{{-{3}}}{{\left({s}+{2}\right)}^{2}}$$
$${y}{\left({s}\right)}=\frac{3}{{s}}+\frac{{-{3}}}{{{s}+{2}}}+\frac{{-{3}}}{{\left({s}+{2}\right)}^{2}}$$
i.e. , $${y}{\left({s}\right)}={L}{\left({y}{\left({t}\right)}\right)}=\frac{3}{{s}}+\frac{{-{3}}}{{{s}+{2}}}+\frac{{-{3}}}{{\left({s}+{2}\right)}^{2}}$$
Step 3
b) Take inverse Laplace transform,
$${L}^{ -{{1}}}{\left({y}{\left({s}\right)}\right)}={L}^{ -{{1}}}{\left(\frac{3}{{s}}+\frac{{-{3}}}{{{s}+{2}}}+\frac{{-{3}}}{{\left({s}+{2}\right)}^{2}}\right)}$$
$$={L}^{ -{{1}}}{\left(\frac{3}{{s}}\right)}-{L}^{ -{{1}}}\frac{{{3}}}{{{s}+{2}}}-{L}^{ -{{1}}}\frac{{{3}}}{{\left({s}+{2}\right)}^{2}}$$
$$={3}{L}^{ -{{1}}}{\left(\frac{3}{{s}}\right)}-{3}{L}^{ -{{1}}}\frac{{{1}}}{{{s}+{2}}}-{3}{L}^{ -{{1}}}\frac{{{1}}}{{\left({s}+{2}\right)}^{2}}$$
$${L}^{ -{{1}}}{\left({y}{\left({s}\right)}\right)}={3}-{3}{e}^{{-{2}{t}}}-{3}{t}{e}^{{-{2}{t}}}$$
Therefore, $${y}{\left({t}\right)}={L}^{ -{{1}}}{\left({y}{\left({s}\right)}\right)}={3}-{3}{e}^{{-{2}{t}}}-{3}{t}{e}^{{-{2}{t}}}$$

### Relevant Questions

Use the table of Laplace transform and properties to obtain the Laplace transform of the following functions. Specify which transform pair or property is used and write in the simplest form.
a) $$x(t)=\cos(3t)$$
b)$$y(t)=t \cos(3t)$$
c) $$z(t)=e^{-2t}\left[t \cos (3t)\right]$$
d) $$x(t)=3 \cos(2t)+5 \sin(8t)$$
e) $$y(t)=t^3+3t^2$$
f) $$z(t)=t^4e^{-2t}$$
Use the Laplace transform to solve the given initial-value problem.
$$dy/dt-y=z,\ y(0)=0$$
Consider the differential equation for a function f(t),
$$tf"(t)+f'(t)-f((t))^2=0$$
a) What is the order of this differential equation?
b) Show that $$f(t)=\frac{1}{t}$$ is a particular solution to this differential equation.
c)Find a particular solution with f(0)=0
2. Find the particular solutions to the differential equations with initial conditions:
a)$$\frac{dy}{dx}=\frac{\ln(x)}{y}$$ with y(1)=2
b)$$\frac{dy}{dx}=e^{4x-y}$$ with y(0)=0
a) Write the sigma notation formula for the right Riemann sum $$R_{n}$$ of the function $$f(x)=4-x^{2}$$ on the interval $$[0,\ 2]$$ using n subintervals of equal length, and calculate the definite integral $$\int_{0}^{2}f(x) dx$$ as the limit of $$R_{n}$$ at $$n\rightarrow\infty$$.
(Reminder: $$\sum_{k=1}^{n}k=n(n+1)/2,\ \sum_{k=1}^{n}k^{2}=n(n+1)(2n+1)/6)$$
b) Use the Fundamental Theorem of Calculus to calculate the derivative of $$F(x)=\int_{e^{-x}}^{x}\ln(t^{2}+1)dt$$
One property of Laplace transform can be expressed in terms of the inverse Laplace transform as $$L^{-1}\left\{\frac{d^nF}{ds^n}\right\}(t)=(-t)^n f(t)$$ where $$f=L^{-1}\left\{F\right\}$$. Use this equation to compute $$L^{-1}\left\{F\right\}$$
$$F(s)=\arctan \frac{23}{s}$$

A random sample of $$n_1 = 14$$ winter days in Denver gave a sample mean pollution index $$x_1 = 43$$.
Previous studies show that $$\sigma_1 = 19$$.
For Englewood (a suburb of Denver), a random sample of $$n_2 = 12$$ winter days gave a sample mean pollution index of $$x_2 = 37$$.
Previous studies show that $$\sigma_2 = 13$$.
Assume the pollution index is normally distributed in both Englewood and Denver.
(a) State the null and alternate hypotheses.
$$H_0:\mu_1=\mu_2.\mu_1>\mu_2$$
$$H_0:\mu_1<\mu_2.\mu_1=\mu_2$$
$$H_0:\mu_1=\mu_2.\mu_1<\mu_2$$
$$H_0:\mu_1=\mu_2.\mu_1\neq\mu_2$$
(b) What sampling distribution will you use? What assumptions are you making? NKS The Student's t. We assume that both population distributions are approximately normal with known standard deviations.
The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations.
The standard normal. We assume that both population distributions are approximately normal with known standard deviations.
The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations.
(c) What is the value of the sample test statistic? Compute the corresponding z or t value as appropriate.
(Test the difference $$\mu_1 - \mu_2$$. Round your answer to two decimal places.) NKS (d) Find (or estimate) the P-value. (Round your answer to four decimal places.)
(e) Based on your answers in parts (i)−(iii), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level \alpha?
At the $$\alpha = 0.01$$ level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
At the $$\alpha = 0.01$$ level, we reject the null hypothesis and conclude the data are statistically significant.
At the $$\alpha = 0.01$$ level, we fail to reject the null hypothesis and conclude the data are statistically significant.
At the $$\alpha = 0.01$$ level, we reject the null hypothesis and conclude the data are not statistically significant.
(f) Interpret your conclusion in the context of the application.
Reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver. (g) Find a 99% confidence interval for
$$\mu_1 - \mu_2$$.
lower limit
upper limit
(h) Explain the meaning of the confidence interval in the context of the problem.
Because the interval contains only positive numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, we can not say that the mean population pollution index for Englewood is different than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains only negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is less than that of Denver.
ALSO, USE PARTIAL FRACTION WHEN YOU ARRIVE
$$L(y) = \left[\frac{w}{(s^2 + a^2)(s^2+w^2)}\right]*b$$
Problem 2 Solve the differential equation
$$\frac{d^2y}{dt^2}+a^2y=b \sin(\omega t)$$ where $$y(0)=0$$
and $$y'(0)=0$$
Use the Laplace transform to solve the following initial value problem:
$$2y"+4y'+17y=3\cos(2t)$$
$$y(0)=y'(0)=0$$
a)take Laplace transform of both sides of the given differntial equation to create corresponding algebraic equation and then solve for $$L\left\{y(t)\right\}$$ b) Express the solution $$y(t)$$ in terms of a convolution integral
Find the Laplace transform of the function $$L\left\{f^{(9)}(t)\right\}$$