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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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    MATLAB solves this ODE in terms of [Biconfluent Heun function](http://dlmf.nist.gov/31.12#E3).2019-05-24

1 Answers 1

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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.