1。(y^4-3x²)dy+xydx=0
解:显然,y=0是原方程的解
于是,设x=ty² (y≠0),则dx=y²dt+2tydy
代入原方程得(y^4-3t²y^4)dy+ty³(y²dt+2tydy)=0
==>(1-t²)dy+tydt=0
==>tdt/(t²-1)=dy/y
==>1/2[1/(t-1)-1/(t+1)]dt=dy/y
==>ln│(t-1)/(t+1)│=2ln│y│+ln│C│ (C是积分常数)
==>(t-1)/(t+1)=Cy²
==>(x-y²)/(x+y²)=Cy²
∴原方程的通解是y=0与(x-y²)/(x+y²)=Cy² (C是积分常数);
2。y³dx+2(x²-xy²)dy=0
解:显然,y=0是原方程的解
于是,设x=ty² (y≠0),则dx=y²dt+2tydy
代入原方程得y³(y²dt+2tydy)+2(t²y^4-ty^4)dy=0
==>ydt+2t²dy=0
==>2dy/y=-dt/t²
==>2ln│y│=1/t+ln│C│ (C是积分常数)
==>y²=Ce^(1/t)
==>y²=Ce^(y²/x)
∴原方程的通解是y=0与y²=Ce^(y²/x) (C是积分常数)。
追问谢谢啊.....但给的答案是:Cy^6=x^2-y^4 (C为积分常数)
y^2=2xln│y│+x (C为积分常数)