First Order ODE/x y' = y + 2 x exp (- y over x)

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Theorem

The first order ordinary differential equation:

$(1): \quad x \dfrac {\d y} {\d x} = y + 2 x e^{-y/x}$


is a homogeneous differential equation with solution:

$e^{y / x} = \ln x^2 + C$


Proof

Let:

$\map M {x, y} = y + 2 x e^{-y/x}$
$\map N {x, y} = x$


Put $t x, t y$ for $x, y$:

\(\ds \map M {t x, t y}\) \(=\) \(\ds t y + 2 t x e^{-t y / t x}\)
\(\ds \) \(=\) \(\ds t \paren {y + 2 x e^{-y / x} }\)
\(\ds \) \(=\) \(\ds t \, \map M {x, y}\)
\(\ds \map N {t x, t y}\) \(=\) \(\ds t x\)
\(\ds \) \(=\) \(\ds t \, \map N {x, y}\)


Thus both $M$ and $N$ are homogeneous functions of degree $1$.

Thus, by definition, $(1)$ is a homogeneous differential equation.


By Solution to Homogeneous Differential Equation, its solution is:

$\ds \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$

where:

$\map f {x, y} = \dfrac {y + 2 x e^{-y / x} } x$


Thus:

\(\ds \ln x\) \(=\) \(\ds \int \frac {\d z} {\dfrac {z + 2 e^{-z} } 1 - z} + C_1\)
\(\ds \) \(=\) \(\ds \int \frac {e^z} 2 \rd z + C_1\)
\(\ds \) \(=\) \(\ds \frac {e^z} 2 + C_1\) Primitive of Exponential Function
\(\ds \leadsto \ \ \) \(\ds 2 \ln x + C\) \(=\) \(\ds e^z\) putting $C = -\dfrac {C_1} 2$
\(\ds \leadsto \ \ \) \(\ds e^{y / x}\) \(=\) \(\ds \ln x^2 + C\)

$\blacksquare$


Sources

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