Question

高等数学求解？

Answer 1

7. 记 p = y', 则 xp' - p +2 = 0, x ≠ 0 时，
p' - p/x = -2/x 是一阶线性微分方程
p = e^(∫dx/x) [ ∫(-2/x)e^(∫-dx/x)dx + 2C ]
= x [ ∫(-2dx/x^2 + 2C ] = x[2/x + 2C] = 2Cx + 2,
即 y' = 2Cx + 2, y = Cx^2 + 2x + D,
x = 0 时， y' = p = 2, y = 2x + D, 已包含在上述通解之中。
曲线过原点， y(0) = 0, 则 D = 0， y = Cx^2 + 2x,
∫<0, 1> (Cx^2 + 2x)dx = [(C/3)x^3 + x^2]<0, 1>
= C/3 + 1 = 2, 得 C = 3， y = 3x^2 + 2x

追问

不好意思 是 11

追答

11. 题目画得太乱了。 右边的积分上限是 x 还是 1 ？ 若是 x：
f'(x) + f(x) = [1/(x+1)] ∫ f(t)dt, f'(0) + f(0) = 0, f'(0) = -f(0) = -1;
(x+1)[f'(x) + f(x)] = ∫ f(t)dt, 两边对 x 求导，得
f'(x) + f(x) + (x+1)[ f''(x) + f'(x)] = f(x)
(x+1)f''(x) + (x+2)f'(x) = 0.
令 p = f'(x), 则 dp/dx = -[(x+2)/(x+1)]p
dp/p = -[(x+2)/(x+1)]dx
lnp = -[x + ln(x+1)] + lnC
f'(x) = p = C/[(x+1)e^x]
f'(0) = C = -1, f'(x) = -1/[(x+1)e^x] = -e^(-x)/(x+1)
f(x) 拟似积不出来。