Question

x-2)dx+4(x+y-1)dy=0

Answer 1

Let x=2+u, y=v-1 then dx=du, dy=dv and x+y-1=u+v.

With this substitution we have a homogenous equation:

udu + 4(u+v)dv = 0

du/dv = -4(1+v/u)

The standard substitution commonly used for this type of ODE is:

s=u/v

Then u=sv, du/dv = s+v ds/dv, v/u = 1/s and the equation will be

s + v ds/dv = -4(1 + 1/s)

This is an equation with separable variables. We write it as follows:

$- \frac{dv}{v} = \frac{ds}{s + 4 +4/s} = \frac {s ds}{(s+2)^2} = (\frac{1}{s+2}-\frac{2}{(s+2)^2})ds = d(\ln|s+2| +\frac{2}{s+2})$

$d(\ln|v(s+2)| +\frac{2}{s+2})=0$

$\ln|v(s+2)| +\frac{2}{s+2}=c$

In the original variables:

$\ln|x+2y| +\frac{2}{2+(x-2)/(y+1)}=c$

$\ln|x+2y| +\frac{2y+2}{2y+x}=c$

$|x+2y| e^{\frac{2y+2}{2y+x}}=e^c$

Further simplification is hardly possible.

The function $(x+2y) e^{\frac{2y+2}{2y+x}}$ is the first integral of the ODE.

In particularly, y=-x/2 is one of the integral curves of this ODE.