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: $$ p = \sqrt{-x \pm \sqrt{x^2 + 2xy^2}} $$
Answer for screen readers
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.
