Linear First Order ODE/y' + (y over x) = k x^n
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Theorem
Let $k, n \in \R$ be real numbers.
- $(1): \quad \dfrac {\d y} {\d x} + \dfrac y x = k x^n$
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:
- $y = \dfrac {k \ln x} x + \dfrac C x$
$\blacksquare$