求微分方程dy/dx=e^(x-2y)的通解,求详解,

Question

求微分方程dy/dx=e^(x-2y)的通解,求详解,

Answer 1

令u=x-2y
则u'=1-2y'
代入原方程得：(1-u')/2=e^u
故u'=1-2e^u
du/(1-2e^u)=dx
d(e^u)*1/[e^u(1-2e^u)]=dx
d(e^u)*[1/e^u+2/(1-2e^u)]=dx
积分：lne^u-ln|1-2e^u)|=x+C1
e^u/|1-2e^u|=Ce^x
e^(x-2y)/|1-2e^(x-2y)|=Ce^x