First Order ODE in form y' = f (a x + b y + c)

Theorem

The first order ODE:

$\dfrac {\d y} {\d x} = \map f {a x + b y + c}$

can be solved by substituting:

$z := a x + b y + c$

to obtain:

$\ds x = \int \frac {\d z} {b \map f z + a}$

Proof

We have:

$\dfrac {\d y} {\d x} = \map f {a x + b y + c}$

Put:

$z := a x + b y + c$

Then:

\(\ds z\)

\(=\)

\(\ds a x + b y + c\)

\(\ds \leadsto \ \ \)

\(\ds \frac {\d z} {\d x}\)

\(=\)

\(\ds a + b \dfrac {\d y} {\d x}\)

\(\ds \leadsto \ \ \)

\(\ds \dfrac {\d y} {\d x}\)

\(=\)

\(\ds \frac 1 b \paren {\frac {\d z} {\d x} - a}\)

\(\ds \leadsto \ \ \)

\(\ds \frac 1 b \paren {\frac {\d z} {\d x} - a}\)

\(=\)

\(\ds \map f z\)

\(\ds \leadsto \ \ \)

\(\ds \frac {\d z} {\d x}\)

\(=\)

\(\ds b \map f z + a\)

This can be solved by Solution to Separable Differential Equation:

$\ds x = \int \frac {\d z} {b \map f z + a}$

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 2.7$: Homogeneous Equations: Problem $3$