超難拉普拉斯1

y" + 2ty' - 4y = 1, y(0) = y'(0) = 0

For L{ty'}(s) = -d/ds L{y'}(s) = -d/ds [sY(s) - y(0)] = -sY'(s) - Y(s)

Apply Laplace Transform on both sides,

L{y" + 2ty' - 4y}(s) = L{1}(s)

[s^2Y(s) - sy(0) - y'(0)] + 2[-sY'(s) - Y(s)] - 4Y(s) = 1/s

-2sY'(s) + (s^2 - 6)Y(s) = 1/s

Y'(s) + (3/s - s/2)Y(s) = -1/2s^2

This is a linear first order equation with integrating factor

u(s) = exp(int(3/s - s/2)ds) = exp(ln(s^3) - s^2/4) = s^3exp(-s^2/4)

d/ds (u(s)Y(s)) = d/ds (s^3exp(-s^2/4)Y(s) = -s/2 exp(-s^2/4)

So, the equation becomes:

s^3 exp(-s^2/4) Y(s) = - int (s/2)exp(-s^2/4) ds = exp(-s^2/4) + C

Y(s) = 1/s^3 + Cexp(s^2/4) / s^3

But lim s→inf Y(s) = 0

So, Y(s) = 1/s^3

and the solution is: y(t) = t^2 / 2

題目的標題寫明的，當然是。

是否必需以拉普拉斯轉換解之?