Solve the differential equation p³ + 2xp - 2xy²/p = 0 in which p = dy/dx.

Understand the Problem

The question asks for the solution of a differential equation given in the form of a problem statement. It specifically refers to solving the equation p³ + 2xp - 2xy²/p = 0 and identifying the relationship between p and y.

Answer

The relationship between $ p $ and $ y $ is given by: $$ p = \sqrt{-x \pm \sqrt{x^2 + 2xy^2}} $$

Steps to Solve

Write the given equation

The given differential equation is

$$ p^3 + 2xp - \frac{2xy^2}{p} = 0 $$

where ( p = \frac{dy}{dx} ).

Multiply through by ( p )

To eliminate the fraction, multiply the entire equation by ( p ):

$$ p^4 + 2xp^2 - 2xy^2 = 0 $$

Rearrange to form a polynomial

Rearranging gives:

$$ p^4 + 2xp^2 - 2xy^2 = 0 $$

This is now a polynomial in ( p ).

Use the quadratic formula

The equation can be treated as a quadratic in ( p^2 ). Let ( u = p^2 ), we can rewrite the equation as:

$$ u^2 + 2xu - 2xy^2 = 0 $$

Applying the quadratic formula ( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ), where ( a = 1, b = 2x, c = -2xy^2 ):

Calculate the discriminant

The discriminant is:

$$ D = b^2 - 4ac = (2x)^2 - 4(1)(-2xy^2) = 4x^2 + 8xy^2 $$

Solve for ( p^2 )

Using the quadratic formula:

$$ p^2 = \frac{-2x \pm \sqrt{4x^2 + 8xy^2}}{2} = -x \pm \sqrt{x^2 + 2xy^2} $$

Thus, ( p = \sqrt{-x \pm \sqrt{x^2 + 2xy^2}} ).

The relationship between ( p ) and ( y ) is given by:

$$ p = \sqrt{-x \pm \sqrt{x^2 + 2xy^2}} $$

More Information

This solution involves finding the relationship between the derivative of ( y ) (denoted as ( p )) and ( x ) and ( y ) through the process of manipulating and solving a polynomial equation derived from the original differential equation.

Tips

Confusing ( p ) with ( p^2 ) when using the quadratic formula.
Forgetting to simplify the discriminant correctly before applying the quadratic formula.

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