Question

微积分问题：求图内两个微分方程的通解

题目如图，要有过程

Answer 1

(c) dx/dy - x = y, 是 x 对 y 的一阶线性微分方程
x = e^(∫dy) [ ∫ye^(-∫dy)dy + C]
= e^y [∫ye^(-y)dy + C] = e^y [-∫yde^(-y) + C]
= e^y [-ye^(-y) + ∫e^(-y)dy + C]
= e^y [-ye^(-y) - e^(-y) + C]
= Ce^y - y - 1
(d) x ≠ 0 时，得 y' + y/x = 2/x
y = e^(-∫dx/x) [ ∫(2/x)e^(∫dx/x)dx + C]
= e^(-lnx)[∫(2/x)e^(lnx)dx + C]
= (1/x) [∫2dx + C] = (2x+C)/x
x = 0 时，y = 2.

Answer 2

(c)
dx/dy = x+y
dx/dy -x =y
xp= Ay + B
dxp/dy = A
dxp/dy -xp =y
-Ay+(-A+B)=y
=>
A=-1 and -A+B=0
A=-1 and B=-1
xp = -y-1
let
xg= Ce^(y)
x= xg+xp
=Ce^(y) - y-1
(d)
x.dy/dx +y=2
d/dx (xy) = 2
xy = 2x + C
y= (2x + C)/x