Solving differential equation y * y '' - (y ') ^ 2-y ^ 2 * y' = 0

Solving differential equation y * y '' - (y ') ^ 2-y ^ 2 * y' = 0

yy''-y'^2=y^2y'
(yy''-y'^2)/y^2=y'
(y'/y)'=y'
Integral on both sides: y '/ y = y + C1
dy/dx=y^2+C1y
dy/[y(y+C1)]=dx
Integral on both sides,
Left = 1 / C1 ∫ (1 / Y-1 / (y + C1)) dy = 1 / c1ln | Y / (y + C1) | + C2
Right = x + C2
So 1 / c1ln | Y / (y + C1) | = x + C2
ln|y/(y+C1)|=C1x+C2
y/(y+C1)=C2e^(C1x)
y=C1C2e^(C1x)/(1-C2e^(C1x))

Elden Ring Premiere

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