求下列微分方程的通解(x^2y^2-1)y'+2xy^3=0
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∵(x^2y^2-1)y'+2xy^3=0
==>(x^2y^2-1)dy+2xy^3dx=0
==>x^2dy-dy/y^2+2xydx=0 (等式两端同除y^2)
==>x^2dy+yd(x^2)+d(1/y)=0
==>d(x^2y)+d(1/y)=0
==>x^2y+1/y=C (C是积分常数)
==>x^2y^2+1=Cy
∴原方程的通解是x^2y^2+1=Cy.
==>(x^2y^2-1)dy+2xy^3dx=0
==>x^2dy-dy/y^2+2xydx=0 (等式两端同除y^2)
==>x^2dy+yd(x^2)+d(1/y)=0
==>d(x^2y)+d(1/y)=0
==>x^2y+1/y=C (C是积分常数)
==>x^2y^2+1=Cy
∴原方程的通解是x^2y^2+1=Cy.
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