Question

高数求通解，求详解！谢谢！

1. dy/dx=(y-x)/(y+x)
2. x^2*dy/dx=xy-y^2

Answer 1

1
dy/dx=(y-x)/(y+x)=(y/x-1)/(y/x+1),
设y=xu，则dy/dx=u+xdu/dx，原方程化为u+du/dx=(u-1)/(u+1),
整理得(u+1)du/(u^2+1)=-dx/x,
两边积分∫u/(u^2+1)du+∫1/(u^2+1)du=-∫1/xdx
1/2(ln(u^2+1))+arctanu=-lnx+lnC1
u=y/x代入上式
ln根号下（y^2/x^2+1）+arctany/x=-lnx+lnC1
ln根号下(y^2+x^2)-ln根号下x^2+lnx-lnC1=-arctany/x
ln(C1*根号下(x^2+y^2))=-arctany/x
C1*根号下(x^2+y^2)=e^(-arctany/x)
根号下(x^2+y^2)=Ce^(-arctany/x)

2
X^2 dy/dx+y^2=xy
dy/dx + y^2 /x^2 = y/x
dy/dx -y/x = -y^2 /x^2
y^(-2) dy/dx - y^(-1) / x = -x^(-2) -----(1)
令y^(-1) = v
(-1) y^(-2) dy/dx = dv/dx
(1) 变为
-dv/dx - v /x = -x^(-2)
dv/dx + v /x = x^(-2) ---------(2)
化为伯努利方程dy/dx + p(x) v = q(x)
其中p(x) = 1/x
q(x) = x^(-2)
积分因子为e^∫p(x) dx = e^∫1/x dx = e^ln(x) = x
(2)式两边同乘x
x (dv/dx + v) = x x^(-2)
d/dx(v x) = x^(-1)
∫ d/dx(v x) = ∫ dx/ x
vx = ln(x) + C
x/y = ln(x) + C
y/x = 1/(ln(x) + C)
y = x / (ln(x) + C)