Question

变量可分离方程 xsecydx+(x+1)dy=0,（1+y)dx-(1-x)dy=0 (xy∧2+x）dx+(1-x∧2）ydy=0.

Answer 1

xsecydx+(x+1)dy=0
dycosy=-xdx/(1+x)
d(siny)=-(1+x)dx/(1+x)+dx/(1+x)
=-dx+d(1+x)/(1+x)
siny=-x+ln(1+x)+c
(1+y)dx-(1-x)dy=0
y=-1,显然是解
y不等于-1,则
dx/(1-x)=dy/(1+y)
-d(1-x)/(1-x)=d(1+y)/(1+y)
-ln(1-x)=ln(1+y)+C
(xy^2+x)dx+(1-x^2)ydy=0
xdx(y^2+1)=(x^2-1)ydy
ydy/(1+y^2)=xdx/(x^2-1)
d(1+y^2)/(1+y^2)=d(x^2-1)/(x^2-1)
ln(1+y^2)=ln(x^2-1)+C
1+y^2=C'(x^2-1)
C'=e^C > 0