# Using Laplace Transform , solve the following differential equation {y}text{}-{4}{y}={e}^{{-{3}{t}}},{y}{left({0}right)}={0},{y}'{left({0}right)}={2} a) frac{14}{{20}}{e}^{{{2}{t}}}-frac{5}{{30}}{e}^{{-{2}{t}}}-frac{9}{{30}}{e}^{{-{6}{t}}} b) frac{11}{{20}}{e}^{{{2}{t}}}-frac{51}{{20}}{e}^{{-{2}{t}}}-frac{4}{{20}}{e}^{{-{3}{t}}} c) frac{14}{{15}}{e}^{{{2}{t}}}-frac{5}{{10}}{e}^{{-{2}{t}}}-frac{9}{{20}}{e}^{{-{3}{t}}} d) frac{14}{{20}}{e}^{{{2}{t}}}+frac{5}{{20}}{e}^{{-{2}{t}}}-frac{9}{{20}}{e}^{{-{3}{t}}}

Question
Laplace transform
Using Laplace Transform , solve the following differential equation
$${y}\text{}-{4}{y}={e}^{{-{3}{t}}},{y}{\left({0}\right)}={0},{y}'{\left({0}\right)}={2}$$
a) $$\frac{14}{{20}}{e}^{{{2}{t}}}-\frac{5}{{30}}{e}^{{-{2}{t}}}-\frac{9}{{30}}{e}^{{-{6}{t}}}$$
b) $$\frac{11}{{20}}{e}^{{{2}{t}}}-\frac{51}{{20}}{e}^{{-{2}{t}}}-\frac{4}{{20}}{e}^{{-{3}{t}}}$$
c) $$\frac{14}{{15}}{e}^{{{2}{t}}}-\frac{5}{{10}}{e}^{{-{2}{t}}}-\frac{9}{{20}}{e}^{{-{3}{t}}}$$
d) $$\frac{14}{{20}}{e}^{{{2}{t}}}+\frac{5}{{20}}{e}^{{-{2}{t}}}-\frac{9}{{20}}{e}^{{-{3}{t}}}$$

2020-12-02
Step 1
As per guidelines, only typed answers are acceptable. so we will give only typed solution.
Given initial value problem is
$${y}\text{}-{4}{y}={e}^{{-{3}{t}}},{y}{\left({0}\right)}={0},{y}'{\left({0}\right)}={2}$$
Step 2
Applying Laplace transform on both sides
$${L}{\left\lbrace{y}\text{}-{4}{y}\right\rbrace}={L}{\left\lbrace{e}^{{-{3}{t}}}\right\rbrace}$$
$$\Rightarrow{L}{\left\lbrace{y}\text{}\right\rbrace}-{4}{L}{\left\lbrace{y}\right\rbrace}=\frac{1}{{{s}+{3}}}$$
$$\Rightarrow{s}^{2}{L}{\left\lbrace{y}\right\rbrace}-{s}{y}{\left({0}\right)}-{y}'{\left({0}\right)}-{4}{L}{\left\lbrace{y}\right\rbrace}=\frac{1}{{{s}+{3}}}$$
$$\Rightarrow{\left({s}^{2}-{4}\right)}{L}{\left\lbrace{y}\right\rbrace}-{2}=\frac{1}{{{s}+{3}}}$$
$$\Rightarrow{L}{\left\lbrace{y}\right\rbrace}=\frac{1}{{{\left({s}+{3}\right)}{\left({s}^{2}-{4}\right)}}}+\frac{2}{{{s}^{2}-{4}}}$$
$$={\left(\frac{1}{{{5}{\left({s}+{3}\right)}}}+\frac{1}{{{20}{\left({s}-{2}\right)}}}-\frac{1}{{{4}{\left({s}+{2}\right)}}}\right)}+{\left(-\frac{1}{{{2}{\left({s}+{2}\right)}}}+\frac{1}{{{2}{\left({s}-{2}\right)}}}\right)}$$
$$=\frac{1}{{{5}{\left({s}+{3}\right)}}}+\frac{11}{{{20}{\left({s}-{2}\right)}}}-\frac{3}{{{4}{\left({s}+{2}\right)}}}$$
Taking inverse Laplace transform on both sides, we get
$${y}={L}^{ -{{1}}}{\left\lbrace\frac{1}{{{5}{\left({s}+{3}\right)}}}+\frac{11}{{{20}{\left({s}-{2}\right)}}}-\frac{3}{{{4}{\left({s}+{2}\right)}}}\right\rbrace}$$
$$={L}^{ -{{1}}}{\left\lbrace\frac{1}{{{5}{\left({s}+{3}\right)}}}\right\rbrace}+{L}^{ -{{1}}}{\left\lbrace\frac{11}{{{20}{\left({s}-{2}\right)}}}\right\rbrace}-{L}^{ -{{1}}}{\left\lbrace\frac{3}{{{4}{\left({s}+{2}\right)}}}\right\rbrace}$$
$$=\frac{1}{{5}}{e}^{{-{3}{t}}}+\frac{11}{{20}}{e}^{{{2}{t}}}-\frac{3}{{4}}{e}^{{-{2}{t}}}$$
Hence, solution is
$${y}{\left({t}\right)}=\frac{11}{{20}}{e}^{{{2}{t}}}-\frac{3}{{4}}{e}^{{-{2}{t}}}+\frac{1}{{5}}{e}^{{-{3}{t}}}\ \text{ or }\ {y}{\left({t}\right)}=\frac{11}{{20}}{e}^{{{2}{t}}}-\frac{15}{{20}}{e}^{{-{2}{t}}}+\frac{4}{{20}}{e}^{{-{3}{t}}}$$
Comparing the answer with options, none of the option is correct.
The option given are all wrong.

### Relevant Questions

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) - ?
Find the Laplace transform $$L\left\{u_3(t)(t^2-5t+6)\right\}$$
$$a) F(s)=e^{-3s}\left(\frac{2}{s^4}-\frac{5}{s^3}+\frac{6}{s^2}\right)$$
$$b) F(s)=e^{-3s}\left(\frac{2}{s^3}-\frac{5}{s^2}+\frac{6}{s}\right)$$
$$c) F(s)=e^{-3s}\frac{2+s}{s^4}$$
$$d) F(s)=e^{-3s}\frac{2+s}{s^3}$$
$$e) F(s)=e^{-3s}\frac{2-11s+30s^2}{s^3}$$
Use Laplace transform to find the solution of the IVP
$$2y'+y=0 , y(0)=-3$$
a) $$f{{\left({t}\right)}}={3}{e}^{{-{2}{t}}}$$
b)$$f{{\left({t}\right)}}={3}{e}^{{\frac{t}{{2}}}}$$
c)$$f{{\left({t}\right)}}={6}{e}^{{{2}{t}}} d) \(f{{\left({t}\right)}}={3}{e}^{{-\frac{t}{{2}}}}$$
The inverse Laplace transform for
$$\displaystyle{F}{\left({s}\right)}=\frac{8}{{{s}+{9}}}-\frac{6}{{{s}^{2}-\sqrt{{3}}}}$$ is
a) $$\displaystyle{8}{e}^{{-{9}{t}}}-{6} \sin{{h}}{{\left({3}{t}\right)}}$$
b) $$\displaystyle{8}{e}^{{-{9}{t}}}-{6} \cos{{h}}{\left({3}{t}\right)}$$
c) $$\displaystyle{8}{e}^{{{9}{t}}}-{6} \sin{{h}}{\left({3}{t}\right)}$$
d) $$\displaystyle{8}{e}^{{{9}{t}}}-{6} \cos{{h}}{\left({3}{t}\right)}$$
Solve by using Laplace Transform and explain the steps in brief
$$\displaystyle\frac{{{d}^{2}{x}}}{{{\left.{d}{t}\right.}^{2}}}+{2}\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}+{x}={3}{t}{e}^{{-{t}}}$$
Given $$x(0)=4, x'(0)=2$$
Given that $$f{{\left({t}\right)}}={4}{e}^{{-{3}{\left({t}-{4}\right)}}}$$
a) Find $${L}{\left[\frac{{{d} f{{\left({t}\right)}}}}{{{\left.{d}{t}\right.}}}\right]}$$ by differentiating f(t) and then using the Laplace transform tables in lecture notes.
b) Find $${L}{\left[\frac{{{d} f{{\left({t}\right)}}}}{{{\left.{d}{t}\right.}}}\right]}$$ using the theorem for differentiation
c) Repeat a) and b) for the case that $$f{{\left({t}\right)}}={4}{e}^{{-{3}{\left({t}-{4}\right)}}}{u}{\left({t}-{4}\right)}$$
The Laplace transform of the function $${\left({2}{t}-{3}\right)}{e}^{{\frac{{{t}+{2}}}{{3}}}}$$ is equal to:
a) $${e}^{{\frac{2}{{3}}}}{\left(\frac{2}{{\left({s}-\frac{1}{{3}}\right)}^{2}}-\frac{3}{{{s}-\frac{1}{{3}}}}\right)}$$
b) $${e}^{{\frac{2}{{3}}}}{\left(\frac{2}{{\left({s}-\frac{1}{{3}}\right)}^{2}}\cdot\frac{3}{{{s}-\frac{1}{{3}}}}\right)}$$
c) $${e}^{{\frac{2}{{3}}}}{\left(\frac{6}{{\left({s}-\frac{1}{{3}}\right)}^{2}}\right)}$$
d) $${\left(\frac{2}{{{\left({s}-\frac{1}{{3}}\right)}^{2}+\frac{2}{{3}}}}\right)}$$
ALSO, USE PARTIAL FRACTION WHEN YOU ARRIVE
$$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$$
Find the Laplace transforms of the following time functions.
Solve problem 1(a) and 1 (b) using the Laplace transform definition i.e. integration. For problem 1(c) and 1(d) you can use the Laplace Transform Tables.
a)$$f(t)=1+2t$$ b)$$f(t) =\sin \omega t \text{Hint: Use Euler’s relationship, } \sin\omega t = \frac{e^(j\omega t)-e^(-j\omega t)}{2j}$$
c)$$f(t)=\sin(2t)+2\cos(2t)+e^{-t}\sin(2t)$$
Gastroenterology
We present data relating protein concentration to pancreatic function as measured by trypsin secretion among patients with cystic fibrosis.
If we do not want to assume normality for these distributions, then what statistical procedure can be used to compare the three groups?
Perform the test mentioned in Problem 12.42 and report a p-value. How do your results compare with a parametric analysis of the data?
Relationship between protein concentration $$(mg/mL)$$ of duodenal secretions to pancreatic function as measured by trypsin secretion:
$$\left[U/\left(k\ \frac{g}{h}r\right)\right]$$
Tapsin secreton [UGA]
$$\leq\ 50$$
$$\begin{array}{|c|c|}\hline \text{Subject number} & \text{Protetion concentration} \\ \hline 1 & 1.7 \\ \hline 2 & 2.0 \\ \hline 3 & 2.0 \\ \hline 4 & 2.2 \\ \hline 5 & 4.0 \\ \hline 6 & 4.0 \\ \hline 7 & 5.0 \\ \hline 8 & 6.7 \\ \hline 9 & 7.8 \\ \hline \end{array}$$
$$51\ -\ 1000$$
$$\begin{array}{|c|c|}\hline \text{Subject number} & \text{Protetion concentration} \\ \hline 1 & 1.4 \\ \hline 2 & 2.4 \\ \hline 3 & 2.4 \\ \hline 4 & 3.3 \\ \hline 5 & 4.4 \\ \hline 6 & 4.7 \\ \hline 7 & 6.7 \\ \hline 8 & 7.9 \\ \hline 9 & 9.5 \\ \hline 10 & 11.7 \\ \hline \end{array}$$
$$>\ 1000$$
$$\begin{array}{|c|c|}\hline \text{Subject number} & \text{Protetion concentration} \\ \hline 1 & 2.9 \\ \hline 2 & 3.8 \\ \hline 3 & 4.4 \\ \hline 4 & 4.7 \\ \hline 5 & 5.5 \\ \hline 6 & 5.6 \\ \hline 7 & 7.4 \\ \hline 8 & 9.4 \\ \hline 9 & 10.3 \\ \hline \end{array}$$
...