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
- 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: $(8)$