Question

求微分方程通解

求微分方程（3x^2+2xy-y^2）dx+(x^2-xy)dy=0的通解。

Answer 1

Answer 2

解：∵(3x²+2xy-y²)dx+(x²-xy)dy=90
同除以x^2
(3+2y/x-(y/x)^2)dx+(1-y/x)dy=0
y/x=u
y=ux y'=u'x+u
(3+2u-u^2)+(1-u)(u'x+u)=0
(3+2u-u^2)/(u-1)-u=u'x
(3+2u-u^2-u^2+u)/(u-1)=u'x
(-2u^2+3u+3)/(u-1)=u'x
(u^2-1.5u-1.5)/(u-1)=-1/2u'x
(u-1)/(u^2-1.5u-1.5)=(u-1)/[(u-0.75)^2-33/16]
＝16(u-1)/[33[16/33(u-0.75)^2]-1]
做到这里自然可设16/33(u-0.75)^2=sec^2t 虽然可做出，但很麻繁，可能是题目错了。
、、
题目改成一下：
解：∵(3x²+2xy-y²)dx+(x²-2xy)dy=0
==>3x²dx+2xydx-y²dx+x²dy-2xydy=0
==>d(x³)+yd(x²)-y²dx+x²dy-xd(y²)=0
==>d(x³)+[yd(x²)+x²dy]-[y²dx+xd(y²)]=0
==>d(x³)+d(x²y)-d(xy²)=0
==>x³+x²y-xy²=C (C是积分常数)

Answer 3