Question

求微积方程.y'=e^(x+y).参考答案是.e^x+e^(-y)+c=0怎么解,

Answer 1

y'=dy/dx=e^(x+y)=e^x*e^y

dy/e^y=e^xdx

e^(-y)dy=e^xdx

-de^(-y)=de^x

则

-e^(-y)=e^x+C

则

e^x+e^(-y)+C=0