Linear First Order ODE/(x^2 + y) dx = x dy

Theorem

The linear first order ODE:

$(1): \quad \paren {x^2 + y} \rd x = x \rd y$

has the general solution:

$y = x^2 + C x$

Proof

Rearranging $(1)$:

$(2): \quad \dfrac {\d y} {\d x} - \dfrac y x = x$

$(2)$ is in the form:

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

where:

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

$\map Q x = x$

Thus:

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

\(=\)

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

\(\ds \)

\(=\)

\(\ds -\ln x\)

\(\ds \)

\(=\)

\(\ds \map \ln {\frac 1 x}\)

\(\ds \leadsto \ \ \)

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

\(=\)

\(\ds \frac 1 x\)

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

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

and the general solution is:

$\dfrac y x = x + C$

or:

$y = x^2 + C x$

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): Miscellaneous Problems for Chapter $2$: Problem $12$