常微分方程求解,重赏![2xy+(x^2)y+(y立方)/3]dx +(x^2 +y^2)dy = 0
2个回答

∵(2xy+x²y+y³/3)dx+(x²+y²)dy=0

==>e^x*(2xy+x²y+y³/3)dx+e^x*(x²+y²)dy=0

==>2xye^xdx+x²ye^xdx+y³e^x/3dx+x²e^xdy+y²e^xdy=0

==>ye^xd(x²)+x²yd(e^x)+y³/3d(e^x)+x²e^xdy+e^xd(y³/3)=0

==>yd(x²e^x)+x²e^xdy+d(y³e^x/3)=0

==>d(x²ye^x)+d(y³e^x/3)=0

∴x²ye^x+y³e^x/3=C (C是积分常数)

故原微分方程的通解是x²ye^x+y³e^x/3=C (C是积分常数).