Question

solve following differential equations:
(equation solvable for p)
p(p-y)=X(X+y)​

Answer 1

The answer for p(p-y)=X(X+y)​ is (2y + x^2 - c )( x + y + 1 - ce^x ) = 0

Given :

p(p-y)=X(X+y)​

Find :

solve of the following p(p-y)=X(X+y)​

Solution :

p ( p - y ) = x ( x + y )

p^2 - yp = x^2  + xy

p^2 - x^2 - yp - xy = 0

( p^2 - x^2 ) - y( p + x ) = 0

( p + x ) ( p - x ) - y( p + x ) = 0

( p + x ). [ ( p - x) - y ] = 0

( p + x ). [ p - x - y ] = 0

first factor equated to zero given

p + x = 0

dy/dx = -x

dy = -xdx

{ 1dy = -{ xdx

y = -x^2/2 + c^1

2y = -x^2 + 2c^1

settibg 2c61 = c

( 2y + x^2 - c )=0....................(eq 1)

second factor equated to zero given

( p - x - y) = 0

dy/dx - y = x        dy/dx + py = 2

p= -1

I.F = e^{ -1 dx

I.F = e^-x

solution of then differential equation

yc e^-x = { x.e^-x dx + c

y e^-x

ye = -x e^-x + e^-x/-1 + c

ye^-x = -xe^-x - e^-x + c

ye^-x + xe^-x + e^-x = c

e^-x( y + x + 1) = c

( x+ y + 1 )/e^x = c

( x + y + 1 ) = ce^x

[ ( x + y + 1) - ce^x ] = 0....................(eq 2)

The final answer is,

(2y + x^2 - c )( x + y + 1 - ce^x ) = 0

Hence, the solution for the differential equation is (2y + x^2 - c )( x + y + 1 - ce^x ) = 0.

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