Question

已知曲线y=f(x)过(0.0)点，且在任意点(x,y)处的切线斜率为x+y求曲线方程

越详细越好，谢谢

Answer 1

y' = x+y
==> dy/dx - y = x
*e^(-x), e^(-x)y' - e^(-x)y = x e^(-x)
e^(-x)y = integral of x e^(-x) dx
y = e^x [ -xe^(-x) - e^(-x) + c] = -x-1 + ce^x
代入 (0,0), c = 1
y = -x-1 + e^x