设f(x)可微,∫〈下限为0,上限为1〉f(tx)dt=(1/2)f(x)+1,当x不等于0时求f(x).

令v = tx,dv = x dt
∫(0,1) ƒ(tx) dt = (1/2)ƒ(x) + 1
∫(0,x) ƒ(v) * 1/x dv = (1/2)ƒ(x) + 1
2∫(0,x) ƒ(v) dv = xƒ(x) + 2x
2ƒ(x) = xƒ'(x) + ƒ(x) + 2
xy' - y = - 2
y' - y/x = - 2/x,IF = ∫ - 1/x dx = e^(- lnx) = 1/x
y'/x - y/x² = - 2/x²
(y/x)' = - 2/x²
y/x = - 2∫ 1/x² dx
y/x = 2/x + c
y = 2 + cx