First Order ODE in form y' = F ((a x + b y + c) over (d x + e y + f)) where a e = b d

Theorem

Formulation 1

The first order ODE:

$(1): \quad \dfrac {\d y} {\d x} = \map F {\dfrac {a x + b y + c} {d x + e y + f} }$

such that:

$a e = b d$

can be solved by substituting:

$z = a x + b y$

to obtain:

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

which can be solved by Solution to Separable Differential Equation.

Formulation 2

The first order ODE:

$(1): \quad \dfrac {\d y} {\d x} = \map F {\dfrac {a x + b y + c} {d x + e y + f} }$

such that:

$a e = b d$

can be solved by substituting:

$z = d x + e y$

to obtain:

$\dfrac {\d z} {\d x} = e \map F {\dfrac {b z + e c} {e z + e f} } + d$

which can be solved by Solution to Separable Differential Equation.

Examples

$\dfrac {\d y} {\d x} = \dfrac {x + y + 4} {x + y - 6}$

The first order ODE:

$(1): \quad \dfrac {\d y} {\d x} = \dfrac {x + y + 4} {x + y - 6}$

has the general solution:

$y - x = 5 \, \map \ln {x + y - 1} + C$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 2.7$: Homogeneous Equations: Problem $4 \text{(b)}$