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