Question

高数题求解，能写下过程吗，感谢

Answer 1

由原式， f(0) = 0+e^0 = 1,
f'(x) = 3f(x)+2e^(2x), f'(x) - 3f(x) = 2e^(2x) 是一阶线性微分方程，
f(x) = e^(∫3dx) [∫2e^(2x)e^(-∫3dx)dx + C]
= e^(3x) [2∫e^(2x)e^(-3x)dx + C]
= e^(3x) [2∫e^(-x)dx + C] = e^(3x) [-2e^(-x) + C]
= -2e^(2x) + Ce^(3x),
f(0) = -2 + C = 1, C = 3, f(x) = -2e^(2x) + 3e^(3x)

Answer 2

解:由(x-1)f(x+1/x-1)+f(x)=x (1) 令y=x+1/x-1, 于是x=y+1/y-1,带入到(1)式得 (2/(y-1))*f(y) + f(y+1/y-1)=y+1/y-1 (2) 将变量y换成x得 (2/(x-1))*f(x) + f(x+1/x-1)=x+1/x-1 (3) 等式两边乘以 x-1,得 2f(x)+(x-1)*f(x+1/x-1)=x+1 (4) 联立(1)，(4)解得 f(x)=1(x不能等于1) 3. f(x)=根号下(x-1) 因为 x^2-x+1>0 恒成立，所以x^2>x-1 |x|>根号下(x-1) 或 根号下(x-1)<|x| 取k=1,即得 |f(x)|<|x| 4. f(x)=x/(x^2+x+1) |f(x)|=|x/(x^2+x+1)|≤k|x| 等价于:|1/(x^2+x+1)|≤k 等价于:x^2+x+1≥1/k x^2+x+1=x^2+x+1/4+3/4=(x+1/2)^2+3/4≥3/4(=1/k) 取k=4/3 则 f(x)≤4/3|x|