Linear First Order ODE/y' + 2 x y = exp -x^2

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Theorem

The linear first order ODE:

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

has the general solution:

$y = \paren {x + C} \map \exp {-x^2}$


Proof

$(1)$ is in the form:

$\dfrac {\d y} {\d x} + \map P x y = \map Q x$

where:

$\map P x = 2 x$
$\map Q x = \map \exp {-x^2}$

Thus:

\(\ds \int \map P x \rd x\) \(=\) \(\ds \int 2 x \rd x\)
\(\ds \) \(=\) \(\ds x^2\)
\(\ds \leadsto \ \ \) \(\ds e^{\int P \rd x}\) \(=\) \(\ds \map \exp {x^2}\)

Thus from Solution by Integrating Factor, $(1)$ can be rewritten as:

$\map {\dfrac {\d} {\d x} } {\map \exp {x^2} y} = \map \exp {-x^2} \map \exp {x^2} = 1$

Hence:

\(\ds \map \exp {x^2} y\) \(=\) \(\ds \int 1 \rd x\)
\(\ds \) \(=\) \(\ds x + C\)
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds \paren {x + C} \map \exp {-x^2}\)

$\blacksquare$


Sources