Question

p²+2xp-3x²=0 where p=Dy/dx​

Answer 1

Given:

p²+2xp-3x²=0 where p=Dy/dx​

To Find:

to find the value of y in terms of x

Step-by-step explanation:

now,

p²+2xp-3x²=0

$p^{2} +3xp - xp -3x^{2} = 0$

p(p+3x) - x(p+3x) = 0

(p+3x)(p-x) = 0

then either p+3x=0 or p-x=0

thus p has two values

p= -3x      and   p= x    ---------(1)

Now according to question p=Dy/dx

putting value in equation 1

p= -3x

$\dfrac{dy}{dx} = -3x$

dy = -3x dx

on intergrating both sides

y =  $-3\dfrac{x^{2} }{2}$

Similarly for other value of p

p = x

$\dfrac{dy}{dx} = x$

dy = xdx

y = $\dfrac{x^{2} }{2}$