∵e^(y^2+x)dx+ydy=0
==>e^(y^2)*e^xdx=-ydy
==>-2ye^(-y^2)dy=2e^xdx
==>e^(-y^2)d(-y^2)=2e^xdx
==>e^(-y^2)=2e^x+C(C是常数)
∴原方程的通解是e^(-y^2)=2e^x+C.