What is the solution for y(pdp)/(dy)-(p)^2+p=0 ?

The solution for y(pdp)/(dy)-(p)^2+p=0 is p=c_{1}y+1

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y(pdp)/(dy)-(p)^2+p=0

y \frac{p d p}{d y} - (p)^{2} + p = 0

p = c_{1} y + 1

Solution steps

y \frac{p d p}{d y} - p^{2} + p = 0

Solve separable ODE:

p = c_{1} y + 1

p = c_{1} y + 1

\frac{d y}{d x} = \frac{2 x + sec ^{2} ( x )}{2 y}, y (0) = - 2

y^{^{'}} = - 5 y e^{t}, y (0) = 1

2 \frac{d y}{d x} = \frac{x e ^{- y}}{x ^{2} + 3}

t^{2} (1 + y^{2}) + 2 y \frac{d y}{d t} = 0

\frac{d P}{d t} = 0.05 P - 0.00125 P (t)^{2}

What is the solution for y(pdp)/(dy)-(p)^2+p=0 ?
The solution for y(pdp)/(dy)-(p)^2+p=0 is p=c_{1}y+1