2yy=1 y的平方

x^2+y^2-2x+y+k=0x^2-2x+1-1+y^2+y+1/4-1/4+k=0(x-1)^2+(y+1/2)^2-5/4+k=0(x-1)^2+(y-1/2)^2=5/4-k5/4-k>0k

∵令y'=p,则y"=pdp/dy代入原方程,得pdp/dy-2yp^3=0==>p(dp/dy-2yp^2)=0∴p=0,或dp/dy-2yp^2=0∵p=0不满足初始条件,舍去∴dp/dy-2yp

既然你的题目是“可降阶的高阶微分方程”,那就应该这样做：再答：

[yy''-(y')^2]/(y^2)=lny(y'/y)'=lnyy'/y=y(lny-1)y'=y^2(lny-1).

由(y'y")'=(y")^2+y'y"及(yy")'=yy"'+y'y"y"(y')^2=[1/3*(y')^3]'代入原方程得：得：(yy")'-y'y"=(y'y")'-y'y"+[1/3*(y

两边同时对y积分得d(yy')=d(0.5y^2(lny-0.5))y'=0.5ylny-1/4y+c1/y积分得y=1/4y^2lny-1/4y^2+C1lny+C2

yy''-y'^2+y'=0x'y'=1y'=1/x'y''=-x''/(x')^2y*(-x''/(x')^2-(1/x')^2+1/x'=0x''y+1-x'=0x''y-x'=-1x''y-x'

yy''=y'^2+y^2y'=dy/dx=py''=dp/dx=(dp/dy)(dy/dx)=pdp/dyypdp/dy=p^2+y^2(y/2)dp^2=p^2dy+y^2dyp^2=uydu/2

这两题都可以化成全微分求解 .点击放大：

2/9再问：过程，谢谢再答：由题目得y/x=2/3xy/xx+yy-yy/xx-yy=y/x-(y/x)²=2/3-4/9=2/9

(4x+3y-1)²+|2x-y+7|=0|2x-y+7|>=0所以只能(4x+3y-1)²=0|2x-y+7|=0即4x+3y-1=02x-y+7=0解得x=-2y=3

原式=[(x²+y²-2y²)/y]/[(x-y)/xy]×1/(x²-y²)=[(x²-y²)/y]×[xy/(x-y)]×1/

xx+yy+4x-6y+13=0整理得：(x+2)^2+(y-3)^2=0那么只有(x+2)=0(y-3)=0x=-2y=3(x^2-2x)/(x^2+3y^2)=(4+4)/(4+3*9)=8/31

yy"-y'^2=y^2y',那么(yy"-y'^2)/y^2=y',注意到y'/y的导数就是(yy"-y'^2)/y^2,所以对等式两边积分得y'/y=y+A(A为常数),那么dy/[(y+A)*y

x^2+x-x^2-y=3x-y=3(x-y)^2=9x^2+y^2-2xy=9(x^2+y^2)/2-xy=9/2

两边同时对y积分得d(yy')=d(0.5y^2(lny-0.5))y'=0.5ylny-1/4y+c1/y积分得y=1/4y^2lny-1/4y^2+C1lny+C2

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这个第一道应该是令y'=p然后y''=dp/dx的x*dp/dx=p*ln(p/x)然后解出p对P积分即可得到答案第二道也是令y'=p的y''=dp/dx*dy/dy=dp/dy*dy/dx=p*dp

x^2-2x+y^2+6y+10=0(x-1)^2+(y+3)^2=0所以x=1,y=-3x+y=-2x^2表示x的平方

x(x+1)-(xx+y)=-3x^2+x-x^2-y=-3x-y=-3(xx+yy)/2-xy=(x^2+y^2-2xy)/2=(x-y)^2/2=(-3)^2/2=9/2再问：是对的吧！再答：当然