unique living asked in 科學數學 · 1 decade ago

Laplace transform to solve ODE


The initial conditions for its function value and its derivative are both zero



1 Answer

  • 1 decade ago
    Favorite Answer

    y"+2ty'-4y=1. y(0)=0, y'(0)=0. L[y"]+2L[ty']-4L[y]=L[1] Let L[y(t)]=Y(s). ==> s^2Y(s)+(-2)d/ds{s*Y(s)}-4Y(s)=1/s ==>(s^2-6)Y(s)-2sY'(s)=1/s ==> Y '(s)+[-s/2+3/s]Y(s)=(-1/2)(1/s^2), a first order linear ode for Y(s), having an integrating factor u(s)= exp[-s^2/4+3lns]=s^3*e^(-s^2/4) ==> Y(s) can be solved by formula: Y(s)=[s^(-3)*e^(s^2/4)]{ e^(-s^2/4)+C} =s^(-3)+(C/2)*s^(-3)e^(s^2/4). C is supposed to be an arbitrary constant, but keeping in mind that Y(s) is the Laplace transform of our solution so limit of Y(s) goes to 0 as t goes to infinity. Therefore C=0, or Y(s)=s^(-3) only. Finally y(t)=L-1{Y(s)}=L-1{1/S^3}=t^2/2.

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