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docs/math/diffeq/1-intro.md

@@ -16,15 +16,15 @@ This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/
 
 \[ a_n(t)y^{(n)}(t) + a_{n-1}(t)+y^{n-1}(t) + \ldots + a_1(t)y'(t) + a_0(t)y(t) = g(t) \]
 
-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*.
+Note that $a_n(t)$ does not depend 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 *solution(s)* to a differential equation over an interval $\alpha < t < \beta$ are any function(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 *integral of validity* for a solution to a differential equation is the largest possible interval containing the initial conditions 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.
 

docs/math/diffeq/2-1st-order.md

@@ -24,7 +24,7 @@ $$
 \mu(t)\frac{dy}{dt} + \mu'(t)y = \mu(t)g(t)
 $$
 
-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.
+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.
 
 \begin{align}
     \int (\mu(t)y(t))' dt &= \int \mu(t)g(t) \\
@@ -88,7 +88,7 @@ Let the following differential equation of the following forms be given.
     \frac{dy}{dx} &= N(y)M(x) \\
 \end{align}.
 
-For the sake of simplicty, select the following form:
+For the sake of simplicity, select the following form:
 
 $$
 N(y) \frac{dy}{dx} = M(x)

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

@@ -14,21 +14,21 @@ $$ ay'' + by' + cy = g(t) $$
 
 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**. In the event that $g(t) = 0$, we say the equation is *homogenous*. Otherwise, the equation is *nonhomogeneous*.
 
 **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:
+Given a second-order homogenous differential equation with constant coefficients, we assume solutions of the following form:
 
 $$ y(t) = e^{rt} $$
 
-Substituting this equation into the differential equationm, we see the following:
+Substituting this equation into the differential equation, 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:
+Thus, we allow the *characteristic equation* of the differential equation to be as follows:
 
 $$ ar^2 + br + c = 0 $$
 
@@ -36,7 +36,7 @@ $$ ar^2 + br + c = 0 $$
 
 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.
+When the two roots to the characteristic equation are discrete roots in the real numbers, we see the following solutions.
 
 $$ y_1(t) = e^{r_1 t} $$
 
@@ -50,7 +50,7 @@ $$ y(t) = c_1 e^{r_1 t} + c_2 e^{r_2 t} $$
 
 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:
+Let the solutions to the characteristic equation be of the following form:
 
 $$ r_{1,2} = \lambda \pm \mu i $$
 
@@ -64,7 +64,7 @@ Recall Euler's Formula:
 
 $$ e^{i \theta} = \cos \theta + i \sin \theta $$
 
-A colliloquy of Euler's formula is the following:
+A corollary of Euler's formula is the following:
 
 $$ e^{-i \theta} = \cos(-\theta) + i \sin(-\theta) = \cos \theta - i \sin \theta $$
 
@@ -83,7 +83,7 @@ $$ y(t) = c_1 e^{\lambda t} \cos(\mu t) + c_2 e^{\lambda t} \sin(\mu t) $$
 
 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.
+Assume the solutions to the characteristic 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
 
@@ -112,7 +112,7 @@ $$
 
 **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
+## Section 3.8 - Nonhomogeneous Differential Equations
 
 This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/NonhomogeneousDE.aspx).
 
@@ -120,11 +120,11 @@ Assume we have the differential equation as follows:
 
 $$ y'' + p(t) y' + q(t) y = g(t) $$
 
-The equivilent homogenous differential equation is
+The equivalent 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.
+**Theorem**. Assume $Y_1(t)$, $Y_2(t)$ are solutions to the nonhomogeneous 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
 
@@ -156,7 +156,7 @@ Assume we have the differential equation as follows:
 
 $$ y'' + p(t) y' + q(t) y = g(t) $$
 
-The equivilent homogenous differential equation is
+The equivalent homogenous differential equation is
 
 $$ y'' + p(t) y' + q(t) y = 0 $$
 

docs/math/diffeq/4-laplace.md

@@ -60,7 +60,7 @@ $$
 
 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
+**Theorem**. Given a function $f(t)$ with $C^n$ continuity, then
 
 $$
 \mathcal{L} \{ f^{(n)} (t) \} = s^n F(s) - s^{n-1} f(0) - s^{n-2} f'(0) - \ldots - s f^{(n-2)} (0) - f^{(n-1)} (0)
@@ -77,6 +77,6 @@ $$
 
 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
+## Section 4.6 - Non-constant Coefficient IVPs
 
 This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/IVPWithNonConstantCoefficient.aspx).