Question

求微分方程y(1+x^2)dy+x(1+y^2)dx=0满足y(1)=1的解

Answer 1

ydy/(1+y^2)=-xdx/(1+x^2)
两边积者睁老分,得1/2ln(1+y^2)=-1/2ln(1+x^2)+C
得(1+y^2)(1+x^2)=C'(C'=e^(2C))
代首升入x=1,y=1得C'=4
所以方程早嫌的解为(1+x^2)(1+y^2)=4