How Can Laplace Transforms Solve an Initial Value Problem?

[diffeqnoob]

Okay, I know this is alot... 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 $$2t^2$$is 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...

$$y(t) = 1 + 3t^2$$