First Order ODE in form y' = F ((a x + b y + c) over (d x + e y + f)) where a e = b d
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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 $4 \text{(b)}$