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Laplace transform to solve an ODE

  1. Nov 23, 2012 #1

    fluidistic

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    Gold Member

    1. The problem statement, all variables and given/known data
    I must solve the following diff. eq. ##tx''-(4t+1)x'+(4t+2)x=0## with the initial condition ##x(0)=0## and the relations ##\mathcal {L }[tx]=-\frac{d \mathcal{L}[x]}{ds}##, ##\mathcal {L} [tx']=-\frac{d [s \mathcal {L}[x]]}{ds}## and ##\mathcal{L}[x']=s \mathcal {L}[x]-x(0)##. Where x is a function of t.


    2. Relevant equations
    Already given.


    3. The attempt at a solution
    Well I've calculated ##\mathcal {L}[x''t]## which gave me ##-s^2 \frac{d \mathcal {L}[x]}{ds}-2s \mathcal {L}[x]##.
    Then I used the linearity of the Laplace transform and I applied the Laplace transform over the equation. Which eventually lead me to ##-\frac{d(\mathcal{L}[x])}{ds} \cdot s^2 + \mathcal{L}[x] (2-3s)=0##.
    This is where I was stuck when starting to write this thread. Because I'm used to obtain an algebraic expression for ##\mathcal {L}[x]##, not a differential equation. I just solved it and reached ##\mathcal{L}[x]=\frac{1}{s^3e^{\frac{2}{s}}}##. Is this the way to go? To finish, I should take the inverse Laplace transform of that.
     
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