How Can Laplace Transforms Solve an Initial Value Problem?

[diffeqnoob]

Okay, I know this is a lot... but I am stuck, so here goes...

Use the method of Laplace transform to solve the initial value problem

y''+3ty'-6y=0, y(0) = 1, y'(0) = 0
L\{y'' + 3ty' - 6y\} = L\{0\}
s^{2}Y(s) - sy(0) - y'(0) + 3L\{ty'\} - 6Y(s) = 0
s^{2}Y(s) - s(1) - 0 - \frac{d}{ds}\left(3 L\{ty'\}\right) -6Y(s) = 0

Now to resolve the - \frac{d}{ds}\left(3 L\{ty'\}\right)
= - \frac{d}{ds}\left(3 L\{ty'\}\right)

= - \frac{d}{ds}3 \left(sY(s) - y(0)\right)

= -3sY'(s) - 3Y(s)


Plugging it back into the eq we now have

s^{2}Y(s) - s - 3sY'(s) - 3Y(s) - 6Y(s) = 0

-3sY'(s) + (s^{2}-9)Y(s) - s = 0

Y'(s) + \left(-\frac{s}{3} + \frac{3}{s}\right)Y(s) = -\frac{1}{3}

\mu = e^{\int\left(-\frac{s}{3} + \frac{3}{s}\right)ds}

\mu = e^{\left(-\frac{s^{2}}{6} + ln(s^{3})\right)

\mu = s^{3} e^{\left(-\frac{s^{2}}{6}\right)

\int\left(\frac{d}{ds}(s^{3}e^{-\left(\frac{s^2}{6}\right)}Y(s)\right) = \int-\left(\frac{1}{3}\right)s^{3}e^{-\left(\frac{s^2}{6}\right)} ds

s^{3}e^{-\left(\frac{s^2}{6}\right)}Y(s) = \int-\left(\frac{1}{3}\right)s^{3}e^{-\left(\frac{s^2}{6}\right)} ds


RIGHT SIDE
=\left(\frac{1}{3}\right)(-3(s^{2}+6)e^{-\left(\frac{s^2}{6}\right)

=(s^2+6)e^{-\left(\frac{s^2}{6}\right) + A

Y(s)=\frac{(s^2+6)}{s^{3}} + \frac{A e^ \frac{s^2{6}{s^{3}}

Limit...as... s \rightarrow \infty...Y(s) = 0...therefore A = 0

Y(s) = \frac{s^2+6}{s^3}



Break down the Inverse Laplace
L^{-1}\{\frac{s^2+6}{s^3}\}

=L^{-1}\{\frac{s^2}{s^3}\} + L^{-1}{\frac{6}{s^3}\}

=L^{-1}\{\frac{1}{s}\} + L^{-1}{\frac{6}{s^3}\}

= 1 + ?


This is where I get lost... I don't know how to do the other side... Please help.

 

[diffeqnoob]

Why does my tex look all ugly and bad ?

 

[Clausius2]

=L^{-1}\{\frac{1}{s}\} + L^{-1}{\frac{6}{s^3}\}

= 1 + ?


This is where I get lost... I don't know how to do the other side... Please help.

What other side? I hope it is not calculating the antitransform of
\frac{6}{s^3} because it's pretty easy, it is in any table.

Hint:L\{t^n\}=\frac{\Gamma(n+1)}{s^{n+1}}

 

[diffeqnoob]

What other side? I hope it is not calculating the antitransform of
\frac{6}{s^3} because it's pretty easy, it is in any table.

Hint:L\{t^n\}=\frac{\Gamma(n+1)}{s^{n+1}}

Yes this is where I was lost because I didn't have that in the table my professor gave me. His said:


L\{t^n\}=\frac{\Gamma(n!)}{s^{n+1}}

I am lost because I only see this problem as:

L\{t^n\}=\frac{\Gamma(2+1)}{s^{2+1}}
= t^2... i know this isn't right.

wait... is this what I think it is? Am I overthinking this. Is the answer:

2t^2 ? A 2 multiplied into the answer would make the numerator a 6. Am I right ?

 

[diffeqnoob]

can someone verify if 2t^2is right ?

 

[diffeqnoob]

What other side? I hope it is not calculating the antitransform of
\frac{6}{s^3} because it's pretty easy, it is in any table.

Hint:L\{t^n\}=\frac{\Gamma(n+1)}{s^{n+1}}

I have looked all over the internet and in my book and I cannot find the above mentioned Laplace Transform formula anywhere. Did you mean (n!) in the numerator vice the (n+1) ? In that case, that would make my answer 3t^2

 

[Clausius2]

For n a positive integer:

\Gamma(n+1)=n!

 

[diffeqnoob]

I appreciate the help

thanks. I can't believe I didn't see that the antitransform was so simple...

y(t) = 1 + 3t^2