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Catchup 1

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# Section 1 - Basic Concepts
## Section 1.1 - Definitions
This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/classes/de/Definitions.aspx).
**Definition**. A *differential equation* is an equation that describes a function in terms of its derivatives. Examples of differential equations include Newton's Laws, among others.
**Definition**. The *order* of a differential equation is the largest derivative present in the equation with a non-zero constant.
**Definition**. A differential equation that only involves derivatives with respect to one variable is called an *ordinary* differential equation (ODE).
**Definition**. A differential equation that describes a function in terms of derivatives with respect to more than one linearly-independent variable is called a *partial* equation.
**Definition**. A *linear* differential equation is any differential equation that cn be written in the following form:
\[ a_n(t)y^{(n)}(t) + a_{n-1}(t)+y^{n-1}(t) + . . . + a_1(t)y'(t) + a_0(t)y(t) = g(t) \]
Note that $a_n(t)$ does not depeond on any derivative of $y$, so the presence of terms such as $e^y$ or $\sqrt{y'}$ signal that the equation is *nonlinear*.
**Definition**. The *solution(s)* to a differential equation over an inverval $\alpha < t < \beta$ are any funcion(s) $y(t)$ that satisfy the differential equation.
**Definition**. The *initial conditions* are a condition or set of conditions that constrain the possible solution sets.
**Definition**. An *Initial Value Problem* is a differential equation along with the appropriate boundary or initial conditions.
**Definition**. The *integral of validity* for a solution to a differential equation is the largest possible interval containing the initial coniditions for which the solution is valid.
**Definition**. The *general solution* to a differential equation is the most general form a solution to a differential equation can take without requiring the initial conditions.
**Definition**. The *actual solution* to a differential equation is the specific solution that satisfies the differential equation and the boundary conditions.
**Definition**. A solution is said to be *explicit* if it can be written in the form $y = y(t)$. Otherwise, it is said to be *implicit*.
## Section 1.2 - Directional Fields
This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/DirectionFields.aspx).
**Definition**. A directional field is the graph of a $t$ vs. $y(t)$, with vectors drawn at each point with a slope corresponding to $y'(t)$. Notably, each arrow will be pointed right (towards increasing $t$).

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

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# Section 3 - Second Order Differential Equations
## Section 3.1 - Basic Concepts
This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/SecondOrderConcepts.aspx).
All second-order differential equations can be written in the following form:
$$ p(t) y'' + q(t) y' + r(t) y = g(t) $$
In the case where $p(t)$, $q(t)$, and $r(t)$ are constants, we write the equation as the following:
$$ ay'' + by' + cy = g(t) $$
This is a second-order differential equation with constant coefficients.
**Definition**. In the event that $g(t) = 0$, we say the equation is *homogenous*. Otherwise, the equation is *nonhomogenous*.
**Definition**. Principal of Superposition. Let $y_1(t)$ and $y_2(t)$ be solutions to a linear, homogenous differential equation. Then, any linear combination of said solutions is also a solution to the differential equation. In other words, with $c_1, c_2 \in \mathbb{R}$, the following is a solution to a differential equation.
$$ y(t) = c_1 y_1(t) + c_2 y_2(t) $$
Given a second-order homogenous differential equation with constant coeffictions, we assume solutions of the following form:
$$ y(t) = e^{rt} $$
Substituting this equation into the differential equationm, we see the following:
$$ e^{rt}(ar^2 + br + c) = 0 $$
Thus, we allow the *charactaristic equation* of the differential equation to be as follows:
$$ ar^2 + br + c = 0 $$
# Section 3.2 - Real & Distinct Roots
This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/RealRoots.aspx).
When the two roots to the charactaristic equation are discrete roots in the real numbers, we see the following solutions.
$$ y_1(t) = e^{r_1 t} $$
$$ y_2(t) = e^{r_2 t} $$
Thus,
$$ y(t) = c_1 e^{r_1 t} + c_2 e^{r_2 t} $$
# Section 3.3 - Complex Roots
This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/ComplexRoots.aspx).
Let the solutions to the charactaristic equation be of the following form:
$$ r_{1,2} = \lambda \pm \mu i $$
Thus, our two solutions are
$$ y_1(t) = e^{(\lambda + \mu i)t} $$
$$ y_2(t) = e^{(\lambda - \mu i)t} $$
Recall Euler's Formula:
$$ e^{i \theta} = \cos \theta + i \sin \theta $$
A colliloquy of Euler's formula is the following:
$$ e^{-i \theta} = \cos(-\theta) + i \sin(-\theta) = \cos \theta - i \sin \theta $$
Thus, we can write our solutions as the following:
\begin{align}
y_1(t) &= e^{(\lambda + \mu i)t} &= e^{\lambda t} e^{i \mu t} &= e^{\lambda t}(\cos(\mu t) + i \sin(\mu t)) \\
y_2(t) &= e^{(\lambda - \mu i)t} &= e^{\lambda t} e^{-i \mu t} &= e^{\lambda t}(\cos(\mu t) - i \sin(\mu t))
\end{align}
A linear combination of the two solutions can be written as the following:
$$ y(t) = c_1 e^{\lambda t} \cos(\mu t) + c_2 e^{\lambda t} \sin(\mu t) $$
## Section 3.4 - Repeated Roots
This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/RepeatedRoots.aspx).
Assume the solutions to the charactaristic equations are $r = r_1 = r_2$. Thus, the two equations $y_t(t)$ and $y_2(t)$ are not linearly independent.
After a *lot* of algebra, we see that
$$y_1(t) = e^{rt}$$
$$y_2(t) = t e^{rt}$$
## Section 3.5 - Reduction of Order
This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/ReductionofOrder.aspx).
Skipped.
## Section 3.6 - Fundamental Set of Solutions, Wronskian
This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/FundamentalSetsofSolutions.aspx).
**Definition**. Given two functions $f(t)$, $g(t)$, the *Wronskian* is defined as
$$
W(f,g) = \det \begin{vmatrix}
f(t) & g(t) \\
f'(t) & g'(t)
\end{vmatrix}
$$
**Definition**. If $W(f, g) \neq 0$, then $f(t)$ and $g(t)$ are said to form a *fundamental set of solutions*, and can be superimposed to form the general solution.
## Section 3.8 - Nonhomogenous Differential Equations
This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/NonhomogeneousDE.aspx).
Assume we have the differential equation as follows:
$$ y'' + p(t) y' + q(t) y = g(t) $$
The equivilent homogenous differential equation is
$$ y'' + p(t) y' + q(t) y = 0 $$
**Theorem**. Assume $Y_1(t)$, $Y_2(t)$ are solutions to the nonhomogenous differential equations. Then, $Y_1(t) - Y_2(t)$ is a solution to the homogenous differential equation. This can be proved by substitution.
Thus, with $y_h(t)$ the solution to the homogenous problem, and $y_p(t)$ the solution to this particular problem, we can say that the general form of the solution to this differential equation is
$$ y(t) = y_h(t) + y_p(t) $$
## Section 3.9 - Undetermined Coefficients
This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx).
We know the following guesses for functions.
| $g(t)$ | $y_p$ guess |
| :- | -: |
| $\alpha e^{\beta t}$ | $A e^{\beta t}$ |
| $a \cos(\beta t)$ | $A \cos(\beta t) + B \sin(\beta t)$ |
| $b \sin(\beta t)$ | $A \cos(\beta t) + B \sin(\beta t)$ |
| $a \cos(\beta t) + \sin(\beta t)$ | $A \cos(\beta t) + B \sin(\beta t)$ |
| n-th degree polynomial | $A_nt^n + A_{n-1}t^{n-1} + A_1 t + A_0$ |
Combine this with the following:
**Theorem**. Given $y_{p_1}(t)$ is a solution to $y'' + p(t)y' + q(t)y = g_1(t)$ and $y_{p_2}(t)$ is a solution to $y'' + p(t)y' + q(t)y = g_2(t)$, then the function $y_{p_1}(t) + y_{p_2}(t)$ is a solution to $y'' + p(t)y' + q(t)y = g_1(t) + g_2(t)$
## Section 3.10 - Variation of Parameters
This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/VariationofParameters.aspx).
Assume we have the differential equation as follows:
$$ y'' + p(t) y' + q(t) y = g(t) $$
The equivilent homogenous differential equation is
$$ y'' + p(t) y' + q(t) y = 0 $$
For this method, we must have $y_1(t)$ and $y_2(t)$ known. Through a lot of math, we see that
$$
y_p = -y_1 \int \frac{y_2(t)g(t)}{W(y_1, y_2)} dt + y_2 \int \frac{y_1(t)g(t)}{W(y_1, y_2)} dt
$$

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# Section 4 - Laplace Transformations
## Section 4.1 - Definition
This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/LaplaceDefinition.aspx).
**Definition**. The *Laplace transform* of a function is given by the following:
$$
\mathcal{L} \{f(t)\}(s) = F(s) = \int_0^{\infty} e^{-st}f(t) dt
$$
## Section 4.2 - Properties
This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/LaplaceTransforms.aspx).
The Laplace Transformation is a linear transformation over functions in $\mathbb{R}[t]$. That is, given $a, b \in \mathbb{R}, f(t), g(t) \in \mathbb{R}[t]$, we know that
$$
\mathcal{L} \{a f(t)\ + b g(t) \}(s) = a F(s) + b G(s)
$$
## Section 4.3 - Inverse Laplace Transformation
This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/InverseTransforms.aspx).
Given $F(s)$, we define the Inverse Laplace Transformation as the following;
$$
f(t) = \mathcal{L}^{-1} \{F(s)\}
$$
## Section 4.4 - Step Function
The step/Heaviside function $u_c(t)$ is defined as 0 if $t < c$, and 1 if $t > c$.
Alternatively, $u(t - c) = H(t - c)$ is 0 if $t < c$, and 1 if $t > c$.
Applying this to the Laplace transform,
$$
\begin{align}
\mathcal{L} \{ u_c(t) f(t-c) \} &= \int_0^{\infty} e^{-st}u_c(t)f(t) dt \\
&= \int_c^{\infty} e^{-st}f(t) dt
\end{align}
$$
If we let $u = t - c$,
$$
\begin{align}
\mathcal{L} \{ u_c(t) f(t-c) \} &= \int_0^{\infty} e^{-s(u+c)}f(u) du \\
&= \int_0^{\infty} e^{-su}e^{-cs}f(u) du \\
&= e^{-cs} \int_0^{\infty} e^{-su}f(u) du \\
&= e^{-cs} F(s)
\end{align}
$$
## Section 4.5 - Laplace Transformation applied to IVPs
This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/IVPWithLaplace.aspx).
**Theorum**. Given a function $f(t)$ with $C^n$ continuity, then
$$
\mathcal{L} \{ f^{(n)} (t) \} = s^n F(s) - s^{n-1} f(0) - s^{n-2} f'(0) - ... - s f^{(n-2)} (0) - f^{(n-1)} (0)
$$
For $n=1, 2$ we see that
$$
\begin{align}
\mathcal{L} \{ y' \} &= sY(s) - y(0) \\
\mathcal{L} \{ y'' \} &= s^2 Y(s) - s y(0) - y'(0)
\end{align}
$$
We can take the Laplace transformation of an IVP, solve for $Y(s)$, then take the inverse to find the solution.
## Section 4.6 - Nonconstant Coefficient IVPs
This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/IVPWithNonConstantCoefficient.aspx).