常微分方程求解,重赏![2xy+(x^2)y+(y立方)/3]dx +(x^2 +y^2)dy = 0[2xy+(x^2)y+(y立方)/3]dx +(x^2 +y^2)dy = 0
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∵(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）.
======以下答案可供参考======
供参考答案1:
由原方程得：
2xydx+(x^2+y^2)dy+(x^2y+y^3/3)dx=0
左边两项为全微分
2xydx+(x^2+y^2)dy=d(x^2y+y^3/3)
于是有d(x^2y+y^3/3)+(x^2y+y^3/3)dx=0
显然x^2y+y^3/3=0是一个特解
若x^2y+y^3/3 ≠0，则
ln|x^2y+y^3/3|+x=C
两解可统一写为
x^2y+y^3/3=Ce^(-x)