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docs/math/diffeq/3-2nd-order.md

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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
+
+$$
+y_p = -y_1 \int \frac{y_2(t)g(t)}{W(y_1, y_2)} dt + y_2 \int \frac{y_1(t)g(t)}{W(y_1, y_2)} dt
+$$