题目

求微分方程的通解（y^4-3x^2)dy+xydx=0

提问时间：2020-10-14

答案

∵（y^4-3x²)dy+xydx=0 ==>[(y^4-3x²)dy+xydx]/y^7=0
==>dy/y³-3x²dy/y^7+xdx/y^6=0
==>-d(1/y²)/2+(x²/2)d(1/y^6)+d(x²/2)/y^6=0
==>-d(1/y²)/2+d(x²/(2y^6))=0
==>-1/y²+x²/(2y^6)=C/2 (C是积分常数)
==>-2y^4+x²=Cy^6
∴原方程的通解是x²-2y^4=Cy^6 (C是积分常数)