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Question-Laplace transform

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,718
Hello!!!! :D
I have to find the solution of the equation [tex] y'(x)+5y(x)=e^{-x} , 0<x<\infty , y(0)=2 [/tex] , using the Laplace transform.
That's what I have done so far:
$$L\{y'(x)+5y(x)\}=L\{e^{-x}\}$$
$$L\{y'(x)\}+5L\{y(x)\}=L\{e^{-x}\}$$
$$sY(s)-y(0)+5Y(s)=\frac{1}{s+1}$$
$$Y(s)=\frac{2s+3}{(s+1)(s+5)}$$

But...which are the restrictions for s?How can I find them? :confused:
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
Re: Question-Laplace trasform

I'm not sure what you mean by restrictions on $s$.

But the next step would be to decompose $Y(s)$ using partial fractions in order to find the inverse Laplace transform of $Y(s)$ (i.e., y(x)).
 

evinda

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MHB Site Helper
Apr 13, 2013
3,718
Re: Question-Laplace trasform

I'm not sure what you mean by restrictions on $s$.

But the next step would be to decompose $Y(s)$ using partial fractions to find the inverse Laplace transform of $Y(s)$ (i.e., y(x)).
Should [tex](s+1)(s+5)[/tex] be greater than [tex]0[/tex]?
 

evinda

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MHB Site Helper
Apr 13, 2013
3,718
Re: Question-Laplace trasform

Should [tex](s+1)(s+5)[/tex] be greater than [tex]0[/tex]?
I found that [tex]Y(s)=\frac{1}{4}\frac{1}{s+1}+\frac{7}{4}\frac{1}{s+5}[/tex].

But,before that,do I have to write what $s $should satisfy?
 

Random Variable

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MHB Math Helper
Jan 31, 2012
253
Re: Question-Laplace trasform

Then you would probably want the real part of $s$ to be greater than $-1$ so that both ILT exist.
 

evinda

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MHB Site Helper
Apr 13, 2013
3,718
Re: Question-Laplace trasform

Then you would probably want the real part of $s$ to be greater than $-1$ so that both ILT exist.
And how can I explain it,why it should be like that? :confused:
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
Re: Question-Laplace trasform

You can tell by where the LT has singularities.

Otherwise notice that the Laplace transform of $\displaystyle e^{-x}$ is $ \displaystyle \frac{1}{s+1}$.

But $ \displaystyle \int_{0}^{\infty} e^{-x} e^{-sx} \ dx = \int_{0}^{\infty} e^{-(1+s) x} \ dx $ converges only when $s > -1 $.


Similarly the Laplace transform of $e^{-5x}$ is $\displaystyle \frac{1}{s+5}$.

But $ \displaystyle \int_{0}^{\infty} e^{-5x} e^{-sx} \ dx = \int_{0}^{\infty} e^{-(5+s) x} \ dx $ converges only when $s > -5 $.


So if you want both transforms to exist at the same time, then $s$ should be greater than $-1$.
 
Last edited:

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,718
Re: Question-Laplace trasform

You can tell by where the ILT has singularities.

Otherwise notice that the Laplace transform of $\displaystyle e^{-x}$ is $ \displaystyle \frac{1}{s+1}$.

But $ \displaystyle \int_{0}^{\infty} e^{-x} e^{-sx} \ dx - \int_{0}^{\infty} e^{-(1+s) x} \ dx $ converges only when $s > -1 $.


Similarly the Laplace transform of $e^{-5x}$ is $\displaystyle \frac{1}{s+5}$.

But $ \displaystyle \int_{0}^{\infty} e^{-5x} e^{-sx} \ dx - \int_{0}^{\infty} e^{-(5+s) x} \ dx $ converges only when $s > -5 $.


So if you want both transforms to exist at the same time, then $s$ (or technically the real part of $s$) should be greater than $-1$.

I understand..thanks for your answer!!And $\int_{0}^{\infty} e^{-(1+s) x} \ dx $ is equal to $\left [\frac{e^{-(s+1)x}}{-(s+1)} \right ]_{0}^{M},M\to \infty=\frac{1}{s+1} $ .Right?
 

Random Variable

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MHB Math Helper
Jan 31, 2012
253
Yes.

And those minus signs in my previous post should be equal signs.
 

evinda

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Apr 13, 2013
3,718

evinda

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MHB Site Helper
Apr 13, 2013
3,718
I have also an other question.Do we conclude that $Y(s)=\frac{1}{4}L\{y_{1}(x)\}+\frac{7}{4}L\{y_{2}(x)\}$ ,or is $\frac{1}{4}L\{y_{1}(x)\}+\frac{7}{4}L\{y_{2}(x)\}$ equal to Y(x) or something else? :confused::confused:
 

evinda

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MHB Site Helper
Apr 13, 2013
3,718
I have also an other question.Do we conclude that $Y(s)=\frac{1}{4}L\{y_{1}(x)\}+\frac{7}{4}L\{y_{2}(x)\}$ ,or is $\frac{1}{4}L\{y_{1}(x)\}+\frac{7}{4}L\{y_{2}(x)\}$ equal to Y(x) or something else? :confused::confused:
Or,is it like that $$Y(s)=\frac{1}{4}L\{y_{1}(x)\}+\frac{7}{4}L\{y_{2}(x)\}$$
$$L^{-1}\{Y(s)\}=L^{-1}\{\frac{1}{4}L\{y_{1}(x)\}+\frac{7}{4}L\{y_{2}(x)\}\}$$ ??
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
I have also an other question.Do we conclude that $Y(s)=\frac{1}{4}L\{y_{1}(x)\}+\frac{7}{4}L\{y_{2}(x)\}$ ,or is $\frac{1}{4}L\{y_{1}(x)\}+\frac{7}{4}L\{y_{2}(x)\}$ equal to Y(x) or something else? :confused::confused:
Or,is it like that $$Y(s)=\frac{1}{4}L\{y_{1}(x)\}+\frac{7}{4}L\{y_{2}(x)\}$$
$$L^{-1}\{Y(s)\}=L^{-1}\{\frac{1}{4}L\{y_{1}(x)\}+\frac{7}{4}L\{y_{2}(x)\}\}$$ ??
Let's pick
$$y(x) = \mathcal L^{-1}\{Y(s)\} = \frac{1}{4}\mathcal L^{-1}\left\{\frac 1 {s+1}\right\}+\frac{7}{4}\mathcal L^{-1}\left\{\frac 1 {s+5}\right\}$$
 

evinda

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MHB Site Helper
Apr 13, 2013
3,718
Let's pick
$$y(x) = \mathcal L^{-1}\{Y(s)\} = \frac{1}{4}\mathcal L^{-1}\left\{\frac 1 {s+1}\right\}+\frac{7}{4}\mathcal L^{-1}\left\{\frac 1 {s+5}\right\}$$
I understand..Thank you very much!!!! :eek: