66 lines
No EOL
1.6 KiB
Markdown
Executable file
66 lines
No EOL
1.6 KiB
Markdown
Executable file
# First-order differential equations
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## First-order linear differential equations
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A first-order **linear** differential equation is one of the type
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$$
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\frac{dy}{dx} + p(x) y = q(x).
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$$
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where $p(x)$ and $q(x)$ are given functions, which may be assumed to be continuous. The equation is called **nonhomogeneous** unless $q(x)$ is dentically zero. The corresponding **homogeneous** equation
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$$
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\frac{dy}{dx} + p(x)y = 0,
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$$
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is separable and so is easily solved. By seperation of variables
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$$
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\frac{dy}{dx} = -p(x)y \implies \int \frac{1}{y}dy = \int p(x)dx.
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$$
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Though, pay attention to absolute values.
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There are two methods for solving nonhomogeneous equations.
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### Integration factor
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The first method is by using an integrating factor. Let $\mu(x)$ be an antiderivative of $p(x)$ and multiply the equation by $e^{\mu(x)}$.
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$$
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\begin{array}{ll}
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e^{\mu(x)} \frac{dy}{dx} + e^{\mu(x)} p(x) y = e^{\mu(x)} q(x) &\implies \frac{d}{dx}(e^{\mu(x)} y) = q(x) e^{\mu(x)}, \\
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&\implies e^{\mu(x)} y = \int q(x) e^{\mu(x)}dx, \\
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&\implies y(x) = e^{-\mu(x)} \int q(x) e^{\mu(x)}dx.
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\end{array}
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$$
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### Variation of the constant
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The second method is by using a variation of a constant. Let $\mu(x)$ be an antiderivative of $p(x)$ and solve the homegeneous equation
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$$
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\frac{dy}{dx} + p(x)y = 0 \implies y(x) = k e^{-\mu(x)}.
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$$
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Try:
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$$
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y'(x) + p(x) y(x) = q(x) = k'(x) e^{\mu(x)},
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$$
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thus $k'(x) = q(x) e^{\mu(x)}$.
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#### Example
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Solve $\frac{dy}{dx} + 2xy = x$ with $y(0) = 3$.
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First solving the homogeneous equation
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$$
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\begin{array}{ll}
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\frac{dy}{dx} + 2xy = 0 &\implies \int \frac{1}{y} dy = -2\int xdx
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\end{array}
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$$ |