Question

1)solve p3-4xyp+8y2=0​

Answer 1

Answer:

Step-by-step explanation:

Given P^3 - 4xyp + 8y^2 = 0 solve the differential equation

Now the given equation can be written as

       4 x = p^2/y + 8y/p

Differentiating with respect to y we get

  4/p = - p^2/y^2 – 2p / y dp/dy + 8/p – 8y/p^2 dp/dy

   -4/p + p^2/y^2 = 2(p/y – 4y/p^2) dp/dy

    P^3 – 4y^2 / py^2 = 2 (p/y – 4y / p^2) dp/dy

                                  = 2. P^3 – 4y^2 / p^2y dp / dy

        Now dy/y = 2 dp/p

     So log c + log y = 2 log p

           p = √cy

Substituting the value of p in the given equation we get

   (cy)^3/2 – 4xy √cy + 8y^2 = 0

hope it helps you!

Answer 2

Answer:

$p^{3} -4x$

Step-by-step explanation: