Find Laplace transform for

$x"-3{x}^{\prime}+2x=1-{e}^{2t}$

e1s2kat26
2021-02-01
Answered

Find Laplace transform for

$x"-3{x}^{\prime}+2x=1-{e}^{2t}$

You can still ask an expert for help

2abehn

Answered 2021-02-02
Author has **88** answers

Step 1

given equation

$x"-3{x}^{\prime}+2x=1-{e}^{2t}$

by assuming that initial condition is zero.

$x"={s}^{2}X(s)$

$\text{and}\text{}{x}^{\prime}=sX(s)$

and laplace transform of$1-{e}^{2t}\text{is}\frac{1}{s}-\frac{1}{s-2}=\frac{s-2-s}{s(s-2)}=\frac{-2}{{s}^{2}-2s}$

Step 2

put the calculated values in the above equation

${s}^{2}X(s)-3sX(s)+2X(s)=\frac{-2}{{s}^{2}-2s}$

$X(s)({s}^{2}-3s+2)=\frac{-2}{{s}^{2}-2s}$

$X(s)=\frac{-2}{({s}^{2}-2s)({s}^{2}-3s+2)}$

$\text{the}X(s)=\frac{-2}{({s}^{2}-2s)({s}^{2}-3s+2)}\text{is the required laplace transform.}$

given equation

by assuming that initial condition is zero.

and laplace transform of

Step 2

put the calculated values in the above equation

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${x}^{\prime}=\frac{t-x}{t},t>0$

I tried to solve it by:

${x}^{\prime}=1-\frac{x}{t}$

$\int dx=\int (1-\frac{x}{t})dt$

$x=t-x\mathrm{ln}t+C$

$x=\frac{t}{1+lnt}+C$

But this doesn't seem right. What am I doing wrong? Also I'm supposed to determine the maximal interval around t=1, but how do I get rid of $C$?

I tried to solve it by:

${x}^{\prime}=1-\frac{x}{t}$

$\int dx=\int (1-\frac{x}{t})dt$

$x=t-x\mathrm{ln}t+C$

$x=\frac{t}{1+lnt}+C$

But this doesn't seem right. What am I doing wrong? Also I'm supposed to determine the maximal interval around t=1, but how do I get rid of $C$?

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Use Laplace transform to solve the initial-value problem

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I have first-order differential equation

$y=xy\prime +\frac{1}{2}{\left(y\prime \right)}^{2}$

Maybe, with this someone will find way to solve it

$\frac{1}{2}{y}^{\prime}(2x+{y}^{\prime})=y$

I thought I can use${x}^{2}+y=t$ for subtitution and when I derivate, I have $t}^{\prime}=2x+{y}^{\prime$ which is acctualy the

$({t}^{\prime}-2x){t}^{\prime}=2t=2{x}^{2}$

same as previous. Who knows?

Maybe, with this someone will find way to solve it

I thought I can use

same as previous. Who knows?