Solution to Linear First Order ODE with Constant Coefficients
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Theorem
A linear first order ODE with constant coefficients in the form:
- $(1): \quad \dfrac {\d y} {\d x} + a y = \map Q x$
has the general solution:
- $\ds y = e^{-a x} \paren {\int e^{a x} \map Q x \rd x + C}$
With Initial Condition
Consider the linear first order ODE with constant coefficients in the form:
- $(1): \quad \dfrac {\d y} {\d x} + a y = \map Q x$
with initial condition $\tuple {x_0, y_0}$
Then $(1)$ has the particular solution:
- $\ds y = e^{-a x} \int_{x_0}^x e^{a \xi} \map Q \xi \rd \xi + y_0 e^{a \paren {x - x_0} }$
Proof 1
From the Product Rule for Derivatives:
\(\ds \map {\frac \d {\d x} } {e^{a x} y}\) | \(=\) | \(\ds e^{a x} \cdot \dfrac {\d y} {\d x} + y \cdot a e^{a x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds e^{a x} \paren {\dfrac {\d y} {\d x} + a y}\) |
Hence, multiplying $(1)$ all through by $e^{\int a \rd x}$:
- $\map {\dfrac \d {\d x} } {e^{a x} y} = e^{a x} \map Q x$
Integrating with respect to $x$ now gives:
- $\ds e^{a x} y = \int e^{a x} \map Q x \rd x + C$
whence we get the result by dividing by $e^{a x}$.
$\blacksquare$
Proof 2
This is a specific instance of Solution to Linear First Order Ordinary Differential Equation:
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:
- $\ds y = e^{-\int P \rd x} \paren {\int Q e^{\int P \rd x} \rd x + C}$
In this instance, we have:
- $\map P x = a$
Hence:
- $\ds \int \map P x = a x$
and the result follows.
$\blacksquare$
Sources
- 1958: G.E.H. Reuter: Elementary Differential Equations & Operators ... (previous) ... (next): Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 1$. The first order equation: $\S 1.2$ The integrating factor