How can I solve xy^2 (p^2+2) =2py^3+x^3? Where p=dy/dx?

Answer 1

We solve this equation as quadratic for p = dy/dx.

We get 2 equatiopns: p = 2y/x - x/y and p = x/y.

First is homoginius and second is separate variable.

Answer 2

How can I solve xy^2 (p^2+2) =2py^3+x^3? Where p=dy/dx?

xy²( p²+2) = 2py³ + x³ , Isolate p, ( p²+2) = 2py³ + x³ / xy² = (2py³ / xy² ) + ( x³/ xy²) = 2 p y/x + x²/y²

Rearrange equation, p²+2 = 2 p y/x + x²/y², then p² - 2 p y/x - ( x²/y² - 2) = 0

This is a quadratic equation in p , find b²-4ac , then b = -2y/x , a = 1, c = -( x²/y² - 2)

b²-4ac = ( -2 y/x)² - 4(1)( -( x²/y² - 2)) = 4 y²/ x² + 4 ( x²/y² - 2) = 4 y²/ x² + 4 x²/y² - 8

p = -b ± √b²-4ac / 2a = 2 y/x ± √ 4 y²/ x² + 4 x²/y² - 8 / 2(1)

simplify p = ( 2 y/x ± 2 √ y²/ x² + x²/y² - 2) / 2 = y/x ± √ y²/ x² + x²/y² - 2

p = dy/dx = y/x ± √ y²/ x² + x²/y² - 2

integrate both sides ∫ dy/dx = ∫ y/x + ∫ √ y²/ x² + x²/y² - 2 or

∫ dy/dx = ∫ y/x - ∫ √ y²/ x² + x²/y² - 2

y = ∫ y/x dx + ∫ (√ y²/ x² + x²/y² - 2 ) dx, and y = y ln x + ∫ (√ y²/ x² + x²/y² - 2 ) dx

y = ∫ y/x dx - ∫ (√ y²/ x² + x²/y² - 2 ) dx, and y = y ln x - ∫ (√ y²/ x² + x²/y² - 2 ) dx