First Order ODE/y dx + (x^2 y - x) dy = 0/Proof 2

Theorem

The first order ODE:

$(1): \quad y \rd x + \paren {x^2 y - x} \rd y = 0$

has the general solution:

$\dfrac {y^2} 2 - \dfrac y x = C$

Proof

Rearranging, we have:

$x^2 y \rd y - \paren {x \rd y - y \rd x} = 0$

Aiming to use Quotient Rule for Differentials, divide by $x^2$:

$y \rd y = \dfrac {x \rd y - y \rd x} {x^2}$

So from Quotient Rule for Differentials: Formulation 1

$y \rd y = \map \d {\dfrac y x}$

from which the solution immediately drops:

$\dfrac {y^2} 2 - \dfrac y x = C$

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 2.9$: Integrating Factors: Example $1$