17,设f(x)为可导函数,且满足∫0到x tf(t)dt=f(x)+x^2 求f(x)

∫[0→x] tƒ(t) dt = ƒ(x) + x²、两边求导
xƒ(x) = ƒ'(x) + 2x
--> xy = y' + 2x
dy/dx = xy - 2x = x(y - 2)
dy/(y - 2) = x dx、两边积分
ln|y - 2| = x²/2 + C'
y - 2 = e^(x²/2 + C') = Ce^(x²/2)
y = 2 + Ce^(x²/2)
即
ƒ(x) = 2 + Ce^(x²/2)、C为任意常数