Solution to Linear First Order ODE with Constant Coefficients/With Initial Condition

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Theorem

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

From Solution to Linear First Order ODE with Constant Coefficients, the general solution to $(1)$ is:

$(2): \quad \ds y = e^{-a x} \int e^{a x} \map Q x \rd x + C e^{-a x}$

Let $y = y_0$ when $x = x_0$.


We have:

$(3): \ds \quad y_0 = e^{-a x_0} \int e^{a x_0} \map Q {x_0} \rd x_0 + C e^{-a x_0}$

Thus:

\(\ds y e^{a x}\) \(=\) \(\ds \int e^{a x} \map Q x \rd x + C\) multiplying $(2)$ by $e^{a x}$
\(\ds y_0 e^{a x_0}\) \(=\) \(\ds \int e^{a x_0} \map Q {x_0} \rd x + C\) multiplying $(3)$ by $e^{a x}$
\(\ds \leadsto \ \ \) \(\ds y e^{a x}\) \(=\) \(\ds y_0 e^{a x_0} + \int e^{a x} \map Q x \rd x - \int e^{a x_0} \map Q {x_0} \rd x\) substituting for $C$ and rearranging
\(\ds \) \(=\) \(\ds y_0 e^{a x_0} + \int_{x_0}^x e^{a \xi} \map Q \xi \rd \xi\) Fundamental Theorem of Calculus
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds e^{a x} \int_{x_0}^x e^{a \xi} \map Q \xi \rd \xi + y_0 e^{-a \paren {x - x_0} }\) dividing by $e^{a x}$ and rearranging

$\blacksquare$


Sources

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