Question

求微分方程sinydy+(cosy-e^x)dx=0的通解

Answer 1

令P=cosy-e^x,Q=siny,

因为aP/ay≠aQ/ax,所以设g=g(x)满足a(Pg)/ay=a(Qg)/ax

即-gsiny=g'siny

dg/g=-dx

一个可行的g为g(x)=e^(-x)

所以e^(-x)sinydy+(e^(-x)cosy-1)dx=0

-e^(-x)d(cosy)-cosyd(e^(-x))=dx

d(e^(-x)cosy)=-dx

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

cosy=e^x*(-x+C)