Linear First Order ODE/y' - (y over x) = k x

Theorem

Let $k \in \R$ be a real number.

The linear first order ODE:

$(1): \quad \dfrac {\d y} {\d x} - \dfrac y x = k x$

has the general solution:

$\dfrac y x = k x + C$

or:

$y = k x^2 + C x$

Proof

$(1)$ is in the form:

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

where $\map P x = -\dfrac 1 x$.

Thus:

\(\ds \int \map P x \rd x\)

\(=\)

\(\ds \int -\frac 1 x \rd x\)

\(\ds \)

\(=\)

\(\ds -\ln x\)

\(\ds \leadsto \ \ \)

\(\ds e^{\int P \rd x}\)

\(=\)

\(\ds e^{-\ln x}\)

\(\ds \)

\(=\)

\(\ds \frac 1 x\)

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

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

and the general solution is:

$\dfrac y x = k x + C$

or:

$y = k x^2 + C x$

$\blacksquare$