148 lines
3.4 KiB
Markdown
148 lines
3.4 KiB
Markdown
# Section 2 - First Order Differential Equations
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## Section 2.1 - Linear Differential Equations
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This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/Linear.aspx).
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Let the following first-order linear differential equation be given, with $p(t)$ and $g(t)$ continuos.
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$$
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\frac{dy}{dt} + p(t)y = g(t)
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$$
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### Deriving the Solution
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Next, we let $\mu(t)$ be our *integrating factor*. Multiply both sides of the equation through by $\mu(t)$.
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$$
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\mu(t)\frac{dy}{dt} + \mu(t)p(t)y = \mu(t)g(t)
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$$
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Now, define $\mu(t)$ so that $\mu(t)p(t) = \mu'(t)$. Thus, we can state the following:
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$$
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\mu(t)\frac{dy}{dt} + \mu'(t)y = \mu(t)g(t)
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$$
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The left of the preceding equation is simply the product rule, so we can write $(\mu(t)y(t))' = \mu(t)g(t)$. Take the integral of both sides.
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\begin{align}
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\int (\mu(t)y(t))' dt &= \int \mu(t)g(t) \\
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\mu(t)y(t) + C &= \int \mu(t)g(t) dt \\
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y(t) &= \frac{\int \mu(t)g(t) dt - C}{\mu(t)}
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\end{align}
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Let $C$ absorb the negative sig, and we see the following.
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$$
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y(t) = \frac{\int \mu(t)g(t) dt + C}{\mu(t)}
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$$
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This is the general solution to the differential equation. However, it is incomplete, as we do not know $\mu(t)$
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To derive the function, recall that we defined $\mu(t)p(t) = \mu'(t)$. Thus, we can rewrite this equation.
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\begin{align}
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\frac{\mu'(t)}{\mu(t)} &= p(t) \\
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(\ln \mu(t))' &= p(t) \\
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\end{align}
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Integrate both sides.
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\begin{align}
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\ln \mu(t) + k = \int p(t) dt \\
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\mu(t) &= e^{\int p(t) dt + k} \\
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&= e^k e^{\int p(t) dt}
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\end{align}
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As $k$ is an unknown constant, rewrite this as $\mu(t) = k \exp(\int p(t) dt)$.
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### Summary
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The following differential equation is given.
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$$
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\frac{dy}{dt} + p(t)y = g(t)
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$$
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To find a solution to this differential equation, construct the **integrating factor**. $\mu(t)$.
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$$\mu(t) = k \exp(\int p(t) dt)$$
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Thus, the solution to the differential equation can be written as the following.
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$$
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y(t) = \frac{\int \mu(t)g(t) dt + C}{\mu(t)}
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$$
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## Section 2.2 - Separable Differential Equations
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This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/Separable.aspx).
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Let the following differential equation of the following forms be given.
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\begin{align}
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N(y) \frac{dy}{dx} &= M(x)
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\frac{dy}{dx} &= \frac{M(x)}{N(y)} \\
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\frac{dy}{dx} &= \frac{N(y)}{M(x)} \\
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\frac{dy}{dx} &= N(y)M(x) \\
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\end{align}.
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For the sake of simplicity, select the following form:
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$$
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N(y) \frac{dy}{dx} = M(x)
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$$
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Thus, integrate both sides with respect to $x$.
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$$
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\int N(y) \frac{dy}{dx} dx = \int M(x) dx
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$$
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Since $y$ is really $y(x)$, we can make the following substitution:
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$$
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u = y(x) \text{ and } du = y'(x)dx = \frac{dy}{dx} dx
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$$
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This reduces the integral to the following:
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$$
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\int N(u) du = \int M(x) dx
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$$
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This is solvable from here.
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## Section 2.4 - Bernoulli Equations
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This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/Bernoulli.aspx).
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Let a differential equation of the following form be given, with $n \in \mathbb{N}; n \geq 2$
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$$
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y' + p(x)y = q(x)y^n
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$$
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This is a *Bernoulli equation*.
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Divide by $y^n$.
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$$
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y^{-n}y' + p(x)y^{1-n} = q(x)
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$$
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Now, make the substitution $v = y^{1-n}$. Thus, the derivative is as follows.
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$$
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v' = (1-n)y^{-n}y'
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$$
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Substituting into the first equation yields the following.
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$$
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\frac{1}{1-n}v' + p(x)v = q(x)
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$$
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After solving, be sure to rewrite in terms of $y$.
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