What are the restrictions for s and how to find them in Laplace transform?

[evinda]

Hello! :D
I have to find the solution of the equation $$y'(x)+5y(x)=e^{-x} , 0<x<\infty , y(0)=2$$ , 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?

 

[polygamma]

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]

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 $$(s+1)(s+5)$$ be greater than $$0$$?

 

[evinda]

Re: Question-Laplace trasform

Should $$(s+1)(s+5)$$ be greater than $$0$$?

I found that $$Y(s)=\frac{1}{4}\frac{1}{s+1}+\frac{7}{4}\frac{1}{s+5}$$.

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

 

[polygamma]

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]

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?

 

[polygamma]

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 $-1$.

 

[evinda]

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?

 

[polygamma]

Yes.

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

 

[evinda]

Yes.

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

Nice..Thank you very much! :)

 

[evinda]

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?

 

[evinda]

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?

Or,is it like that $$Y(s)=\frac{1}{4}L\{y_{1}(x)\}+\frac{7}{4}L\{y_{2}(x)\}$$
$$L^{-1}\{Y(s)\}=L^{-1}\{\frac{1}{4}L\{y_{1}(x)\}+\frac{7}{4}L\{y_{2}(x)\}\}$$ ??

 

[I like Serena]

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?

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]

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!