Question

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

Answer 1

Answer:

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

Answer 2

Answer:

Explanation:

Solve the equation p

3

– 4xyp + 8y

2

= 0.