Solution to Linear First Order Ordinary Differential Equation/Proof 1
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Theorem
A linear first order ordinary differential equation in the form:
- $\dfrac {\d y} {\d x} + \map P x y = \map Q x$
has the general solution:
- $\ds y = e^{-\int P \rd x} \paren {\int Q e^{\int P \rd x} \rd x + C}$
Proof
Consider the first order ordinary differential equation:
- $\map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$
We can put our equation:
- $(1): \quad \dfrac {\d y} {\d x} + \map P x y = \map Q x$
into this format by identifying:
- $\map M {x, y} \equiv \map P x y - \map Q x, \map N {x, y} \equiv 1$
We see that:
- $\dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x} = \map P x$
and hence:
- $\map P x = \dfrac {\dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x} } N$
is a function of $x$ only.
It immediately follows from Integrating Factor for First Order ODE that:
- $e^{\int \map P x \rd x}$
is an integrating factor for $(1)$.
So, multiplying $(1)$ by this factor:
- $e^{\int \map P x \rd x} \dfrac {\d y} {\d x} + e^{\int \map P x \rd x} \map P x y = e^{\int \map P x \rd x} \map Q x$
The result follows by an application of Solution to Exact Differential Equation.
$\blacksquare$
Sources
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 2.10$: Linear Equations: Problem $1$