First Order ODE/x dy = (x^5 + x^3 y^2 + y) dx

Theorem

The first order ODE:

$(1): \quad x \rd y = \paren {x^5 + x^3 y^2 + y} \rd x$

has the general solution:

$\arctan \dfrac x y = -\dfrac {x^4} 4 + C$

Proof

Rearranging, we have:

$y \rd x - x \rd y = -\paren {x^2 + y^2} x^3 \rd x$

from which:

$\dfrac {y \rd x - x \rd y} {x^2 + y^2} = - x^3 \rd x$

From Differential of Arctangent of Quotient:

$\map \d {\arctan \dfrac x y} = \dfrac {y \rd x - x \rd y} {x^2 + y^2}$

from which $(1)$ evolves into:

$\map \d {\arctan \dfrac x y} = -x^3 \rd x$

Hence the result:

$\arctan \dfrac x y = -\dfrac {x^4} 4 + C$

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 2.9$: Integrating Factors: Problem $4 \ \text{(c)}$