First Order ODE/(exp y - 2 x y) y' = y^2
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Theorem
The first order ODE:
- $(1): \quad \paren {e^y - 2 x y} y' = y^2$
has the general solution:
- $x y^2 = e^y + C$
Proof
Let $(1)$ be rearranged as:
- $\dfrac {\d y} {\d x} = \dfrac {y^2} {e^y - 2 x y}$
Hence:
- $(2): \quad \dfrac {\d x} {\d y} + \dfrac 2 y x = \dfrac {e^y} {y^2}$
It can be seen that $(2)$ is a linear first order ODE in the form:
- $\dfrac {\d x} {\d y} + \map P y x = \map Q y$
where:
- $\map P y = \dfrac 2 y$
- $\map Q y = \dfrac {e^y} {y^2}$
Thus:
\(\ds \int \map P y \rd y\) | \(=\) | \(\ds \int \dfrac 2 y \rd y\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 \ln y\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \ln y^2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds e^{\int P \rd y}\) | \(=\) | \(\ds y^2\) |
Thus from Solution by Integrating Factor, $(2)$ can be rewritten as:
- $\map {\dfrac \d {\d y} } {x y^2} = e^y$
and the general solution is:
- $x y^2 = e^y + C$
$\blacksquare$
Sources
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 2.10$: Problem $4 \ \text{(a)}$