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If I have an equation I want to solve, such as $tx''+t^2x'-3x=0$ using $x(0)=0$, how can I easily reason what the Laplace of the terms multiplied by $t$ would be?

Can I do the following:

$\mathcal{L}\{x''(t)\} = s^2X(s)+x'(0)$

$\mathcal{L}\{x'(t)\} = sX(s)$

So we have:

$-[s^2X(s)+x'(0)]' + [sX(s)]'' -3X(s) = 0$

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    [Here](http://math.stackexchange.com/questions/257553/use-laplace-transform-to-find-a-solution-for-txxtx-0) you have reasoned quite well.2012-12-13
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    @Artem is my reasoning correct above? Also, is it possible to remove the $x'(0)$ term, or must it be given as a condition in order for that to be possible?2012-12-13
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    No, it is not correct.2012-12-13
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    MATLAB solves this ODE in terms of [Biconfluent Heun function](http://dlmf.nist.gov/31.12#E3).2012-12-13

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You need to use that $$ \mathcal L\{x''(t)\}=s^2X(s)-sx(0)-x'(0),\\ \mathcal L\{x'(t)\}=sX(s)-x(0). $$ The initial conditions should be given.

You should also use the fact that $$ \mathcal L\{t^n f(t)\}=(-1)^n\frac{d^n}{ds^n}F(s), $$ if $\mathcal L\{f(t)\}=F(s)$.

So, for you example, you would get $$ -(s^2X-x'(0))'+(sX)''-3X=\\ =-2sX-s^2X'+2X'+sX''-3X=\\ sX''+(2-s^2)X'+(3-2s)X=0, $$ which is not any better than the original one.