x+yy'=2y solve it in separable method

Question

Differential equation of first order

Answer 1

x+yy'=2y, y'=2-x/y.

Let y=vx, y'=v+xv', v+xv'=2-1/v, v²+xvv'=2v-1, (v-1)²=-xvv'.

Separating the variables:

vdv/dx/(v-1)²=-dx/x, (v/(v-1)²)dv=-dx/x,

Let u=v-1, du=dv, v=u+1 so ∫(v/(v-1)²)dv=∫[(u+1)/u²]du=∫du/u+∫du/u².

So, integrating:

ln|u|-1/u=ln|A/x|, where A is a constant,

ln|v-1|-1/(v-1)=ln(A/x), substituting u=v-1,

ln|y/x-1|-1/(y/x-1)=ln(A/x), substituting v=y/x, rewriting:

ln|(y-x)/x|-x/(y-x)=ln(A/x), ln|y-x|-x/(y-x)=ln(A)=C.

CHECK

ln|y-x|-x/(y-x)=C.

Differentiating:

(1/(y-x))(y'-1)-(y-xy')/(y-x)²=0,

y'-1-(y-xy')/(y-x)=0,

(y'-1)(y-x)-y+xy'=0,

yy'-y+x-y=0, x+yy'=2y.