Question

yp^2+(x-y)p-x=0 solve the following differential equations
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Answer 1

Answer:

yP^2 + ( x - y ) P - x = 0

Step-by-step explanation:

discriminant = b²-4ac

D = ( x - y )^2 - 4(y)(-x)

D = x^2 + y^2 -

D = x^2 + y^2 + 2xy

D = ( x + y )^2

$discriminant \: = \: {b}^{2} - 4ac$

$root \: = \frac{ - b \: + - \: \sqrt{d} }{2a}$

$= \frac{ - (x - y) \: + - \sqrt{ {(x + y)}^{2} } }{2y}$

$= \frac{ - x + y + x + y}{2y} \: \: \: \: \: \: \: \: \: \: \frac{ - x + y - x - y}{2y}$

$= \frac{2y}{2y} \: \: \: \: \: \: \: \: \: \: \frac{ - 2x}{ \: \: 2y}$

1 , -(x/y)

First root is 1

Second root is -x/y