Question

(y - px) (p-1)=p obtain its general and singular solution

Answer 1

$(y - px)(p - 1) = p \\ \\ yp - y - p^2x + px = p \\ \\ yp - y - p^2x + px - p = 0 \\ - (yp + y + {p}^{2}x - px + p) = 0$

Answer 2

Answer:

Simplifying

(y + -1px)(p + -1) = p

Reorder the terms:

(-1px + y)(p + -1) = p

Reorder the terms:

(-1px + y)(-1 + p) = p

Multiply (-1px + y) * (-1 + p)

(-1px * (-1 + p) + y(-1 + p)) = p

((-1 * -1px + p * -1px) + y(-1 + p)) = p

((1px + -1p2x) + y(-1 + p)) = p

(1px + -1p2x + (-1 * y + p * y)) = p

Reorder the terms:

(1px + -1p2x + (py + -1y)) = p

(1px + -1p2x + (py + -1y)) = p

Reorder the terms:

(1px + py + -1p2x + -1y) = p

(1px + py + -1p2x + -1y) = p

Solving

1px + py + -1p2x + -1y = p

Solving for variable 'p'.

Reorder the terms:

-1p + 1px + py + -1p2x + -1y = p + -1p

Combine like terms: p + -1p = 0

-1p + 1px + py + -1p2x + -1y = 0

The solution to this equation could not be determined.

Step-by-step explanation: