Question

xy'+y=xy^2lnx 请用伯努利方程解

Answer 1

x(1/y^2)dy+(1/y)dx=xlnxdx

[-xd(1/y)+(1/y)dx]/x^2=[(lnx)/x]dx

-d[1/(xy)]=(1/2)d(lnx)^2

-1/(xy)=(1/2)(lnx)^2+c