Question

微分方程的问题4xy+2x+(2x^2+3y^2)y'=04xy+2yx^2+(2x^2+3y^2)y'=02cos(x

Answer 1

1

4xy+2x+(2x^2+3y^2)y'=0

yd(2x^2)+2x^2dy+dx^2+dy^3=0

2x^y+x^2+y^3=C

2

4xy+2yx^2+(2x^2+3y^2)y'=0

4xydx+2yx^2dx+2x^2dy+3y^2dy=0

3

2cos(x+y)-2xsin(x+y)-2xsin(x+y)y'=0

dx/x=-dcos(x+y)/cos(x+y)

lnx=C-lncos(x+y)

4

1/x+y+(3y^2+x)y'=0

dx/x+d(xy+3y^2)=0

lnx+xy+y^3=C

5

y+e^x+xy'=0

d(xy+e^x)=0

xy+e^x=C

6

3y^4-1+12xy^3y'=0

3y^4dx-dx+xd(3y^4)=0

3xy^4-x=C