求微分方程的通解 x^2*y*dx-(x^3+y^4)dy=0

2024-11-18 10:21:45
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回答(1):

解:∵x^2*ydx-(x^3+y^4)dy=0
==>x^2dx/y^3-x^3dy/y^4-dy=0 (等式两端同除y^4)
==>3(x^2dx/y^3-x^3dy/y^4)-3dy=0
==>d(x^3/y^3)-3dy=0
==>∫d(x^3/y^3)-3∫dy=0
==>x^3/y^3-3y=C (C是积分常数)
==>x^3=Cy^3+3y^4
∴此方程的通解是x^3=Cy^3+3y^4。