First Order ODE/x dy - y dx = (1 + y^2) dy
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Theorem
The first order ODE:
- $(1): \quad x \rd y - y \rd x = \paren {1 + y^2} \rd y$
has the general solution:
- $\dfrac x y = \dfrac 1 y - y + C$
Proof
Rearranging, we have:
- $\dfrac {y \rd x - x \rd y} {y^2} = -\paren {\dfrac 1 {y^2} + 1} \rd y$
From the Quotient Rule for Derivatives:
- $\map \d {\dfrac x y} = \dfrac {y \rd x - x \rd y} {y^2}$
from which:
- $\map \d {\dfrac x y} = -\paren {\dfrac 1 {y^2} + 1} \rd y$
Hence the result:
- $\dfrac x y = \dfrac 1 y - y + C$
$\blacksquare$
Sources
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 2.9$: Integrating Factors: Problem $4 \ \text{(a)}$