First Order ODE/y dx - x dy = x y^3 dy
Jump to navigation
Jump to search
Theorem
The first order ODE:
- $(1): \quad y rd x - x \rd y = x y^3 \rd y$
has the general solution:
- $\ln \dfrac x y = \dfrac {y^3} 3 + C$
Proof
Rearranging, we have:
- $\dfrac {y \rd x - x \rd y} {x y} = y^2 \rd y$
From Differential of Logarithm of Quotient:
- $\map \d {\ln \dfrac y x} = \dfrac {y \rd x - x \rd y} {x y}$
from which:
- $\map \d {\ln \dfrac x y} = y^2 \rd y$
Hence the result:
- $\ln \dfrac x y = \dfrac {y^3} 3 + C$
$\blacksquare$
Sources
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 2.9$: Integrating Factors: Problem $4 \ \text{(b)}$