Solve ${y}^{\u2033}+2{y}^{\prime}+y=\{\begin{array}{ll}4{e}^{t},& 0\le t<1\\ 0,& t\ge 1\end{array}$ where $y(0)=0,{y}^{\prime}(0)=0$

hommequidort0h
2022-09-19
Answered

Solve ${y}^{\u2033}+2{y}^{\prime}+y=\{\begin{array}{ll}4{e}^{t},& 0\le t<1\\ 0,& t\ge 1\end{array}$ where $y(0)=0,{y}^{\prime}(0)=0$

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asked 2022-09-20

Find inverse laplace transform of $f(s)=\frac{1}{(s-2{)}^{2}+9}$

Here is what I have gotten from partial fractions. I observed that ${s}^{2}-4s+13$ is irreducible (doesn't have real roots).

$$\frac{1}{(s-2{)}^{2}+9}=\frac{1}{{s}^{2}-4s+13}=\frac{As+B}{{s}^{2}-4s+13}$$

A=0, B=1

The corresponding value in my table is ${e}^{at}\mathrm{sin}(bt)$ for the corresponding laplace transform $\frac{b}{(s-a{)}^{2}+{b}^{2}}$

so plugging in A and B, I get ${e}^{0t}\mathrm{sin}(1\cdot t)=\mathrm{sin}(t)$. However my textbook answer is $\frac{1}{3}{e}^{2t}\mathrm{sin}3t$

Here is what I have gotten from partial fractions. I observed that ${s}^{2}-4s+13$ is irreducible (doesn't have real roots).

$$\frac{1}{(s-2{)}^{2}+9}=\frac{1}{{s}^{2}-4s+13}=\frac{As+B}{{s}^{2}-4s+13}$$

A=0, B=1

The corresponding value in my table is ${e}^{at}\mathrm{sin}(bt)$ for the corresponding laplace transform $\frac{b}{(s-a{)}^{2}+{b}^{2}}$

so plugging in A and B, I get ${e}^{0t}\mathrm{sin}(1\cdot t)=\mathrm{sin}(t)$. However my textbook answer is $\frac{1}{3}{e}^{2t}\mathrm{sin}3t$

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If the Laplace Transforms of fimetions ${y}_{1}(t)={\int}_{0}^{\mathrm{\infty}}{e}^{-st}{t}^{3}dt,{y}_{2}(t)={\int}_{0}^{\mathrm{\infty}}{e}^{-st}\mathrm{sin}2tdt,{y}_{3}(t)={\int}_{0}^{\mathrm{\infty}}{e}^{-st}{e}^{t}{t}^{2}dt$ exist ,
then Which of the following is the value of $L\{{y}_{1}(t)+{y}_{2}(t)+{y}_{3}(t)\}$

$a)\frac{3}{({s}^{4})}+\frac{2}{({s}^{2}+4)}+\frac{2}{(s-1{)}^{3}}$

$B)\frac{(3!)}{({s}^{3})}+\frac{s}{({s}^{2}+4)}+\frac{(2!)}{(s-1{)}^{3}}$

$c)\frac{3!}{({s}^{4})}+\frac{2}{({s}^{2}+2)}+\frac{1}{(s-1{)}^{3}}$

$d)\frac{3!}{({s}^{4})}+\frac{4}{({s}^{2}+4)}+\frac{2}{({s}^{3})}\cdot \frac{1}{(s-1)}$

$e)\frac{3!}{({s}^{4})}+\frac{2}{({s}^{2}+4)}+\frac{2}{(s-1{)}^{3}}$

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