F(x)=Int(e^(-t^2)cosxt dt) 0 to inf following the leibnitz rule (http://en.wikip

Question

F(x)=Int(e^(-t^2)cosxt dt) 0 to inf
following the leibnitz rule (http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign)
dF/dx=Int((d(e^(t^2)cosxt)/dx)dt) 0 to inf (differentiation inside the integral is partial differentiation with respect to x)
=Int(-xe^(t^2)sinxt dt) 0 to inf
again,
d2F/dx2 = Int(-x^2 e^(t^2)cosxt dt) 0 to inf
=-x^2 Int(e^(t^2)cosxt dt) 0 to inf (since integration is with respect to t, x is a constant)
=-x^2 F
So, we get the differential equation in terms of x as
d2F/dx2+x^2F=0
Honestly, i have no clue about how to solve this diff eq :| (Thot i cud be of some help to provide the procedure this far :P )

I got this a solution from a previous question I asked. Anyone knows how to solve the differential equation?

Explanation / Answer

So, we get the differential equation in terms of x as d2F/dx2+x^2F=0 d2F/dx2 = -x^2F (1/F)d2F = -x^2dx2 integrating, lnF dF = (-1/3)x^3 dx + C1 Again integrating, F * lnF - F = (-1/12)x^4 + C1x + C2

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