A solution for Poisson's differential equation (not Laplace that has a zero constant) in 2D?

I mean the one having a non-zero constant please, and Thanks :)

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  • hfshaw
    Lv 7
    8 years ago
    Favorite Answer

    The solution to any partial differential equation depends on the boundary and initial conditions. Your question makes no sense.

  • 4 years ago

    tx'' + (5t ? 6)x' + 5x = 0. enable F(s) = L{x(t)}. **observe that L{t * f} = - (d/ds) L{f} for any f. Taking the Laplace rework on the two facets: L{t x'' + (5t ? 6)x' + 5x} = L{0} ==> L{tx''} + 5 L{tx'} ? 6 L{x'} + 5 X(s) = 0 ==> - (d/ds)[s^2 F(s) - s x(0) - x'(0)] - 5 (d/ds)[s F(s) - x(0)] ? 6 [s F(s) - x(0)] + 5 F(s) = 0 ==> - (d/ds)[s^2 F(s) - x'(0)] - 5 (d/ds)[s F(s)] ? 6 [s F(s)] + 5 F(s) = 0, on account that x(0) = 0. ==> - [2s F(s) + s^2 F'(s)] - 5 [F(s) + s F'(s)] ? 6 [s F(s)] + 5 F(s) = 0 ==> (s^2 + 5s) F'(s) + 8s F(s) = 0 ==> F' / F = -8/(s + 5) So, ln F = -8 ln(s + 5) + A ==> F(s) = B(s + 5)^(-8), the place B = e^A. Inverting yields x(t) = B e^(-5t) * t^7 / 7! = Ct^7 * e^(-5t), the place C = B/7!. Now, we can discover C with the aid of ability of x(a million) = e^(-5). x(a million) = e^(-5) = C * a million * e^(-5) ==> C = a million. for that reason, x(t) = t^7 * e^(-5t). i'm hoping this allows!

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