Linear First Order ODE/y' + (y over x) = k x^n

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Theorem

Let $k, n \in \R$ be real numbers.

The linear first order ODE:

$(1): \quad \dfrac {\d y} {\d x} + \dfrac y x = k x^n$

has the general solution:

$y = \begin{cases} \dfrac {k x^{n + 1} } {n + 2} + \dfrac C x & : n \ne -2 \\ & \\ \dfrac {k \ln x} x + \dfrac C x & : n = -2 \end{cases}$


Proof

$(1)$ is in the form:

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

where:

$\map P x = \dfrac 1 x$
$\map Q x = k x^n$

Thus:

\(\ds \int \map P x \rd x\) \(=\) \(\ds \int \frac 1 x \rd x\)
\(\ds \) \(=\) \(\ds \ln x\)
\(\ds \leadsto \ \ \) \(\ds e^{\int P \rd x}\) \(=\) \(\ds e^{\ln x}\)
\(\ds \) \(=\) \(\ds x\)

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

$\map {\dfrac \d {\d x} } {x y} = k x^{n + 1}$

When $n + 1 \ne -1$, Primitive of Power is used to obtain:

$x y = \dfrac {k x^{n + 2} } {n + 2} + C$

or:

$y = \dfrac {k x^{n + 1} } {n + 2} + \dfrac C x$


When $n + 1 = -1$, we have:

$\map {\dfrac \d {\d x} } {x y} = \dfrac k x$

and Primitive of Reciprocal is used, yielding:

$x y = k \ln x + C$

or:

$y = \dfrac {k \ln x} x + \dfrac C x$

$\blacksquare$