Question

t*f(t)-∫f(x)dx(此处积分的下限为0,上限为t）=1/2*t2*f(t),求f(t).

Answer 1

没错,不过两边求导后就变成一条微分方程了,要解关于x、y和y'的方程

x·f(x) - ∫(0→x) f(t) dt = (1/2)x²f(x)、两边对x求导

f(x) + xf'(x) - f(x) = (1/2)[2xf(x) + x²f'(x)]

xf'(x) = xf(x) + (1/2)x²f'(x)、即

2y' = 2y + xy'

y'·(2 - x) = 2y

y'/y = 2/(2 - x)

(lny)' = 2/(2 - x)、两边取积分

lny = 2∫ dx/(2 - x) = - 2[ln(2 - x) + lnC₁] = ln[(C₁(2 - x))^(- 2)] = ln[1/(C₁(2 - x))²]

y = 1/[C₁²(2 - x)]² = C₂/(2 - x)²

即f(x) = C₂/(2 - x)²、其中C₁、C₂都是任意常数