Solve this differential equation

following equation is this :

(y - y^2 - x^2)dx - xdy = 0

y–y²−x²= xy'

−1 − (x/y)² = xy'/y² − 1/y = d(−x/y)/dx

Put z = x/y so DE becomes −1−z² = −dz/dx → dz/(1+z²) = dx

Integrate : tan‾¹ (z) = x+c → z=tan(x+c) → y = xcot(x+c)

1) Rearranging the given one, dy/dx = (y - y² - x²)/x

2) Let y = vx; differentiating both sides with respect to x, dy/dx = (dv/dx)x + v

==> (y - y² - x²)/x = (vx - v²x² - x²)/x = v - v²x - x

3) Hence, substituting these in (1), we get,

(dv/dx)x + v = v - v²x - x

==> (dv/dx)x = -x(1 + v²)

Separating the above, -dv/(1+v²) = dx

4) Integrating the above, cot⁻¹(v) = x + C

==> v = cot(x + C)

==> (y/x) = cot(x + C)

==> y = x*cot(x + C)

[Alternatively, in integrating at step 4, it can also be tan⁻¹(v) = -x + C
==> Answer as y = x*tan(C-x)]

(y - y^2 - x^2)dx - xdy = 0

Re-arrange

(y - y^2 - x^2)dx = xdy

dy/dx = 1/x*(y - y^2 -x^2)
dy/dx -y/x +y^2/x = -x

Make a guess that y = ax^n and substitute

nax^n-1 -ax^n-1 +a^2x^2n-1 = -x

If n = 1 then

a - a +a^2 x = -x

a^2 = -1

a = i = sqrt(-1)

So y = ix

Check
(y - y^2 - x^2)dx - xdy = 0

(ix +x^2 -x^2)dx -idx = 0

(y - y^2 - x^2) dx - x dy = 0

dy/dx = (y - y^2 - x^2)/x

dy/dx = (y/x) - (y^2/x) - x -------------------(1)

let y/x = v

y = vx
dy/dx = v + x dv/dx

substitite dy/dx = v + x dv/dx, y = vx and y^2 = v^2 x^2 in eqn (1)

v + x d v/dx = v + v^2 x - x

subtract v from both sides
x dv /dx = v^2 x - x

divide both sides by x
dv/dx = v^2 - 1

separating variables

dv / (v+1)(v-1) = dx

1/2 [ 1/(v+1) + 1/(v-1) ] dv = dx

ln(v + 1) + ln (v - 1) = 2x`+ c

=> ln (v^2 - 1) = 2x + c

substitute back v = y/x

ln(y^2/x^2 - 1) = 2x + c

( y^2 /x^2) - 1 = Ce^(2x)

y^2 / x^2 = 1 + Ce^(2x)

y^2 = x^2 [ 1 + C e^(2x) ]

y = x sqrt[ 1 + C e^(2x) ]

ydx-xdy=x^2dx+y^2dx
ydx-xdy=(x^2+y^2)dx
dividing by y^2 on both sides
[ydx-xdy]/y^2 = d(x/y) =(x/y)^2dx+dx= [(x/y)^2+1]dx

now on integrating
x/y = (x/y)^2+1+c