Monsters exist, but they are too few in number to be truly dangerous. More dangerous are the common men, the functionaries ready to believe and to act without asking questions, Primo Levi

Life is a math equation. In order to gain the most, you have to know how to convert negatives into positives, Anonymous

An algebraic equation is a mathematical statement that declares or asserts the equality of two algebraic expressions. These expressions are constructed using:

**Dependent and independent variables**. Variables represent unknown quantities. The independent variable is chosen freely, while the dependent variable changes in response to the independent variable.**Constants**. Fixed numerical values that do not change.**Algebraic operations**. Operations such as addition, subtraction, multiplication, division, exponentiation, and root extraction.

Definition. A differential equation is an equation that involves one or more dependent variables, their derivatives with respect to one or more independent variables, and the independent variables themselves, e.g., $\frac{dy}{dx} = 3x +5y, y’ + y = 4xcos(2x), \frac{dy}{dx} = x^2y+y, etc.$

It involves (e.g., $\frac{dy}{dx} = 3x +5y$):

**Dependent variables**:*Variables that depend on one or more other variables*(y).**Independent variables**: Variables upon which the dependent variables depend (x).**Derivatives**: Rates at which the dependent variables change with respect to the independent variables, $\frac{dy}{dx}$

The Existence and Uniqueness Theorem provides crucial insight into the behavior of solutions to first-order differential equations ODEs. It states that if:

- The function f(x, y) (the right-hand side of the ODE) in y’ = f(x, y) is continuous in a neighborhood around a point (x
_{0}, y_{0}) and - Its partial derivative with respect to y, $\frac{∂f}{∂y}$, is also continuous near (x
_{0}, y_{0}).

Then the differential equation y' = f(x, y) has a unique solution to the initial value problem through the point (x_{0}, y_{0}) .

A first-order linear differential equation (ODE) has the general form: a(x)y' + b(x)y = c(x) where y′ is the derivative of y with respect to x, and a(x), b(x), and c(x) are functions of x. If c(x) = 0, the equation is called homogeneous, i.e., a(x)y’ + b(x)y = 0.

The equation can also be written in the standard linear form as: y’ + p(x)y = q(x) where $p(x)=\frac{b(x)}{a(x)}\text{ and }q(x) = \frac{c(x)}{a(x)}$

A second-order linear homogeneous differential equation ODE with constant coefficients is a differential equation of the form: y'' + Ay' + By = 0 where:

- y is the dependent variable (a function of the independent variable t),
- y′ and y′′ are the first and second derivatives of y with respect to t,
- t is the independent variable,
- A and B are constants.

This equation is homogeneous, meaning that there are no external forcing terms (like a function of t) on the right-hand side.

The Laplace Transform of a function f(t), where t ≥ 0, is defined as $\mathcal{L}(f(t)) = \int_{0}^{∞} f(t)e^{-st}dt = F(s)$.

One of the most important properties of the Laplace Transform is linearity, which states: $\mathcal{L}(af(t)+bg(t)) = a\mathcal{L}(f(t))+b\mathcal{L}(g(t))$

Function | Laplace Transform |
---|---|

u(t) | $\mathcal{L}(u(t)) = \frac{1}{s}, s > 0$ |

$e^{at}$ | $\frac{1}{s - a}, s > a$ |

$e^{(a + bi)t}$ | $\mathcal{L}(e^{(a + bi)t}) = \frac{1}{s - (a + bi)}, s > a$ |

$\cos(\omega t)$ | $\mathcal{L}(\cos(\omega t)) = \frac{s}{s^2 + \omega^2}, s > 0$ |

$\sin(\omega t)$ | $\mathcal{L}(\sin(\omega t)) = \frac{\omega}{s^2 + \omega^2}, s > 0$ |

$t^n$ | $\mathcal{L}(t^n) = \frac{n!}{s^{n+1}}, s > 0$ |

$u(t-a)$ | $\mathcal{L}(u(t-a)) = \frac{e^{-as}}{s}, s > 0$ |

$\delta(t-a)$ | $\mathcal{L}(\delta(t-a)) = e^{-as}, a \geq 0$ |

$\frac{1}{t}$ | $\mathcal{L}\left(\frac{1}{t}\right) = \text{not defined}$ |

$e^{-bt} \cos(\omega t)$ | $\mathcal{L}(e^{-bt} \cos(\omega t)) = \frac{s + b}{(s + b)^2 + \omega^2}, s > -b$ |

$e^{-bt} \sin(\omega t)$ | $\mathcal{L}(e^{-bt} \sin(\omega t)) = \frac{\omega}{(s + b)^2 + \omega^2}, s > -b$ |

$e^{at}f(t)$ | $\mathcal{L}(e^{at}f(t)) = F(s-a)$ |

This is the Exponential Shift Theorem, indicating that multiplying a function by an exponential term shifts its Laplace Transform.

Besides, $\mathcal{L}(f’(t)) = sF(s)-f(0), \mathcal{L}(f’’(t)) = s^2F(s)-sf(0)-f’(0)$

The Laplace Transform is a powerful mathematical tool for solving linear ordinary differential equations (ODEs), particularly those with constant coefficients and initial value problems (IVPs). By transforming differential equations in the time domain into algebraic equations in the complex frequency domain, we can simplify the process of finding solutions to these equations.

Consider the following second-order linear ODE with constant coefficients: y’’ + Ay’ + By = h(t) where:

- y(t) is the unknown function of time we want to solve for.
- A and B are constants.
- h(t) is a given forcing function (which could be zero in the homogeneous case).
- The initial conditions are y(0) = y
_{0}and y’(0) = y’_{0}

Our goal is to solve this equation for y(t) using the Laplace Transform. The key idea of using the Laplace Transform is that **it converts the time-domain differential equation into an algebraic equation in the complex frequency variable s**. This transformation simplifies solving the differential equation by turning it into an algebraic problem.

First, let’s recall how the Laplace Transform operates on functions and their derivatives.

**Laplace Transform of a function y(t):**$\mathcal{L}(y(t)) = \int_{0}^{∞} e^{-st}y(t)dt = Y(s)$**Laplace Transform of the first derivative y′(t)**: $\mathcal{L}(y’(t)) = sY(s) -y(0) = sY(s) -y_0$**Laplace Transform of the first derivative y’′(t)**: $\mathcal{L}(y’(t)) = s^2Y(s) -sy(0) -y’(0) = s^2Y(s) -sy_0 -y’_0.$

These formulas are derived using integration by parts and take into account the initial conditions.

Suppose y(t) is a solution to the previous differential equation, $y(t) \leadsto Y(s)$. $\mathcal{L}(\text{I.V.P}) = \text{Algebraic equation in } \mathcal{Y}$.

Applying the Laplace Transform to each term in the differential equation: $\mathcal{L}(y’’(t)+Ay’(t)+By(t)) = \mathcal{L}{h(t)} ↭[\text{Substituting the Laplace Transforms:}] (s^2Y(s) -sy_0 -y’_0) + A(sY(s) -y_0) + BY(s) = H(s)$ where $H(s) = \mathcal{L}({h(t)})$

Next, collect all terms involving Y(s) on one side: $Y(s)(s^2+As+B) = H(s) +sy_0 + y’_0 +Ay_0$

Then, solve for Y(s): $Y(s) = \frac{H(s) +sy_0 + y’_0 +Ay_0}{s^2+As+B}$

This equation represents the solution in the frequency domain, represented as Y(s), an algebraic expression in terms of s. Typically, the result is a rational function $\frac{p(s)}{q(s)}$, where p(s) and q(s) are polynomials in S.

To find the solution y(t) in the time domain, we need to apply the Inverse Laplace Transform to Y(s): $y(t) = \mathcal{L}^{-1}(Y(s))$.

This step may involve:

**Partial Fraction Decomposition**: If Y(s) is a rational function (a ratio of polynomials), decompose it into simpler fractions whose inverse Laplace Transforms are known.**Use of Inverse Transform Formulas**: Utilize standard Laplace Transform tables or formulas to find the inverse transforms of each term. $\mathcal{L}^{-1}(\frac{1}{s-a}) = e^{at}, {L}^{-1}(\frac{1}{s^2+a^2}) = sin(at), {L}^{-1}(\frac{s}{s^2+a^2}) = cos(at)$**Sum the results**. Combine the inverse transforms of all the terms to obtain y(t).

Conclusion: By transforming the differential equation into an algebraic equation using the Laplace Transform, we simplify the problem of solving for y(t). This method is particularly useful for initial value problems and can handle non-homogeneous equations with various forcing functions h(t).

Let f(t) be a function for which we want to calculate the Laplace Transform of its derivative f′(t), $\mathcal{L}(f’(t))$, in terms of the Laplace Transform of the original function.

The Laplace Transform of the derivative f’(t) is defined as: $\mathcal{L}(f’(t)) = \int_{0}^{∞} e^{-st}f’(t)dt$

To simplify the integral, we use integration by parts ∫udv = uv − ∫vdu. We choose:

- Let u = e
^{-st}, so that du = -se^{-st} - Let dv = f’(t)dt, v = f(t).

The formula for integration by parts is ∫udv = uv − ∫vdu, we get:

$e^{-st}f(t)\bigg|_{0}^{∞}$ [Part I]

$-\int_{0}^{∞} -se^{-st}f(t)dt$ [Part II]

The first term of the integration by parts is: $\lim_{t \to ∞} e^{-st}f(t)-e^0f(0)$

We need to calculate: $\lim_{t \to ∞} e^{-st}f(t) = \lim_{t \to ∞} \frac{f(t)}{e^{st}} = 0$

**For this term to vanish as t → ∞, f should be of exponential type**, meaning that: $|f(t)| < ce^{kt}$ for some constants c and k. Under this assumption, as long as s > k, the factor e^{-st} decays to zero faster than f(t) grows. Hence, the first term of the integration by parts [Part I] = $0 -e^0f(0) = -f(0)$

[Part II] The remaining integral is: $-\int_{0}^{∞} -se^{-st}f(t)dt$

$\mathcal{L}(f’(t)) = \int_{0}^{∞} e^{-st}f’(t)dt = -f(0) + s\int_{0}^{∞} e^{-st}f(t)dt = -f(0) + sF(s)$

Assuming that f is of exponential type, the Laplace Transform of the first derivative of f is $\mathcal{L}(f’(t)) = sF(s)-f(0)$

We can extend this process to higher-order derivatives.

How do we know the Laplace Transform of the second derivate? Well, a second derivative is just a first derivate, isn’t it? f’’(t) = [f’(t)]'

$\mathcal{L}(f’’(t)) = s\mathcal{L}(f’(t))-f’(0) = s[sF(s)-f(0)] -f’(0) = s^2F(s) -sf(0)-f’(0)$

By applying this process repeatedly, we can generalize the result for the n-th derivative, f^{(n)}(t). The Laplace Transform of the n-th derivative is given by: $\mathcal{L}(f^{(n)}(t)) = s^nF(s)-s^{n-1}f(0) -s^{n-2}f’(0)-···-f^{n-1}(0)$

Let’s solve the second-order linear differential equation y’’ -y = e^{-t}, with initial conditions y(0) = 1, y’(0) = 0.

Typically, solving such equations involves finding the general solution to the homogeneous equation y′′ −y = 0, calculating the characteristic equation, finding its roots, determining a particular solution, and applying the initial conditions. However, we will use the Laplace Transform method to streamline the process.

**Step 1: Take the Laplace Transform of Both Sides**

Applying the Laplace Transform to each term: y’’ -y =e^{-t}, we get $\mathcal{L}(y’’)-\mathcal{L}(y) = \mathcal{L}(\mathcal{e^{-t}})$ [For simplicity, we use capital Y instead of F], $\mathcal{L}(\mathcal{e^{-t}}) = \frac{1}{s+1}$.

**Step 2: Substitute into the Differential Equation**

Hence, y’’ -y =e^{-t} $\leadsto (s^2Y(s) -s·1 -0) -Y(s) = \frac{1}{s+1}↭ s^2Y(s)-s -Y(s) = \frac{1}{s+1} ⇒ (s^2-1)Y = \frac{1}{s+1}+s = \frac{1 + s^2+s}{s+1}$

**Step 3: Solve for Y(s)**

$Y = \frac{s^2+s+1}{(s^2-1)(s+1)} = \frac{s^2+s+1}{(s-1)(s+1)(s+1)} = \frac{s^2+s+1}{(s-1)(s+1)^2}$

**Step 4: Partial Fraction Decomposition**

$\frac{s^2+s+1}{(s-1)(s+1)^2} = \frac{A}{(s+1)^2} + \frac{B}{s+1} + \frac{C}{s-1} =$

To determine the constants A, B, and C, we multiply both sides by (s−1)(s+1)^{2} and match coefficients of powers of s: s^{2} +s +1 = A(s−1) +B(s+1) +C(s+1)^{2} ↭[Expanding terms:] s^{2} +s + 1 = A(s−1)+B(s+1)(s-1)+C(s+1)^2 = A(s−1)+B(s^{2}-1)+C(s^{2} +2s +1) =[Group terms] = (B+C)s^{2} +(A+2C)s +(−A-B+C)⇒[By comparing coefficients, we get the following system of equations:]

$\begin{cases} B + C = 1 \\ A + 2C = 1 \\ −A -B + C = 1 \end{cases}$

Solving this system yields: A = ^{-1}⁄_{2}, B = ^{1}⁄_{4}, and C = ^{3}⁄_{4}.

**Step 5: Write the Partial Fraction Decomposition**

Thus, the partial fraction decomposition is:

$Y(s) = \frac{\frac{-1}{2}}{(s+1)^2} + \frac{\frac{1}{4}}{s+1} + \frac{\frac{3}{4}}{s-1}$

**Step 6. Apply the Inverse Laplace Transform** to each term:
$\frac{\frac{-1}{2}}{(s+1)^2} + \frac{\frac{1}{4}}{s+1} + \frac{\frac{3}{4}}{s-1} \leadsto \mathcal{L}^{-1}~ \frac{-1}{2}te^{-t} + \frac{1}{4}e^{-t} + \frac{3}{4}e^t$, and the below paragraph explains the first term.

Laplace Transform of Polynomial: $t \leadsto \frac{1!}{s^{1+1}} = \frac{1}{s^2}$. The Exponential Shift Formula (f(t) = t, $e^{at}f(t) \leadsto F(s-a)$): $e^{-1·t}t \leadsto \frac{1}{(s+1)^2} ↭ te^{-t} \leadsto \frac{1}{(s+1)^2}$. Besides, $e^{a·t} \leadsto \frac{1}{s-a}$

Therefore, the solution is $\frac{-1}{2}te^{-t} + \frac{1}{4}e^{-t} + \frac{3}{4}e^t$ where

- The first term, $\frac{-1}{2}te^{-t}$, arises from the non-homogeneous term e
^{-t}. - The second and third terms $\frac{1}{4}e^{-t}, \frac{3}{4}e^t$ form the complementary (homogeneous) solution to the equation y′′−y=0, which comes from the characteristic equation of the homogeneous part. The constants were determined to match the initial conditions y(0)=1 and y′(0)=0.

Initial conditions y(0) = 1, y’(0) = 0.

**Step 1: Apply Laplace Transform**: $\mathcal{L}(y’’) -3·\mathcal{L}(y’) + 2Y(s) = 0$

**Step 2: Substitute into the Differential Equation**:

Using the transforms of the derivatives: (s^{2}Y(s) −sy(0) −y′(0)) -3(sY(s)-y(0)) +2Y(s)=0 ↭[Substitute the initial conditions] (s^{2}Y(s) -s·1 -0) + -3(sY(s)-1) + 2Y(s) = 0 ↭ s^{2}Y(s) -s -3sY(s) + 2Y(s) = 0 ↭[Collect like terms] (s^{2}-3s +2)Y(s) + (-s +3) = 0

**Step 3: Solve for Y(s)**: $Y(s) = \frac{s-3}{s^2-3s+2} =[\text{Factor the denominator:}] \frac{s-3}{(s-1)(s-2)}$

**Step 4: Partial Fraction Decomposition**: $\frac{s-3}{(s-1)(s-2)} = \frac{A}{s-1}+\frac{B}{s-2} = \frac{A(s-2)+B(s-1)}{(s-1)(s-2)} = \frac{(A+B)s-2A-B}{(s-1)(s-2)}$

$\begin{cases} A + B = 1 \\ -2A -B = -3 \end{cases}$

Adding both equations: -A = -2 ⇒A = 2. A + B = 1 ⇒B = 1 -A = 1 -2 = -1.

$Y(s) = \frac{2}{s-1}+\frac{-2}{s-2}$

**Step 4: Apply the Inverse Laplace Transform**

Use standard transform, $c$. Therefore, $y(t)= 2e^{t}-2e^{2t}$

Suppose we have the following IVP: y’’+ 4y = 0 with y(0) = 1, y’(0) = 0.

**Step 1: Take the Laplace Transform of Both Sides**: $\mathcal{L}(y’’) + 4·\mathcal{L}(y) = 0$

**Step 2: Substitute into the Differential Equation**:

Using the transforms of the derivatives: (s^{2}Y(s) −sy(0) −y′(0)) +4Y(s)=0 ↭[Substitute the initial conditions] (s^{2}Y(s) -s·1 -0) + 4Y(s) = 0 ↭ (s^{2} +4)Y(s) = s$

**Step 3: Solve for Y(s)**: $Y(s)=\frac{s}{s^2+4}$

**Step 4: Apply the Inverse Laplace Transform**

Using a table of transforms, we know: $\mathcal{L}^{-1}(\frac{s}{s^2+a^2}) = cos(at)$. Here, a = 2, so the solution to the IVP is: y(t) = cos(2t).

Suppose we have the following IVP:y^{(4)} -y = 0 with y(0) = 0, y’(0) = 1, y’’(0) = 0, y’’’(0) = 0

**Step 1: Take the Laplace Transform of Both Sides**: $\mathcal{L}(y^{(4)}) - \mathcal{L}(y) = 0$

**Step 2: Substitute into the Differential Equation**:

Using the transforms of the derivatives: (s^{4}Y(s) −s^{3}y(0) −s^{2}y′(0) -sy’’(0)-y’’’(0)) - Y(s)=0 ↭[Substitute the initial conditions] (s^{4}Y(s) -s^{2}·1) -Y(s) = 0 ↭ s^{4}Y(s) -s^{2} -Y(s) = 0.

**Step 3: Solve for Y(s)**: $Y(s)=\frac{s^2}{s^4-1}$

**Step 4. Partial Fraction Decomposition**: $Y(s)=\frac{s^2}{s^4-1} = [\text{Factor the denominator}] \frac{s^2}{(s^2-1)(s^2+1)} = \frac{s^2}{(s-1)(s+1)(s^2+1)} = \frac{A}{s-1} + \frac{B}{s+1} + \frac{Cs + D}{s^2+1}$

$s^2 = A(s+1)(s^2+1) + B(s-1)(s^2+1) + (Cs+D)(s-1)(s+1)$

s = -1 ⇒ 1 = B(-4)⇒ B = $\frac{-1}{4}$. s = 1 ⇒1 = A(4) ⇒ A = $\frac{1}{4}$. s = 0 ⇒ $0 = \frac{1}{4}+\frac{1}{4}-D ⇒ D = \frac{1}{2}$. s = 2⇒$4 = \frac{1}{4}·15 + \frac{-1}{4}·5 + (C·2 +\frac{1}{2})·3⇒ 4 = \frac{10}{4} + 6C + \frac{3}{2} ⇒16 = 10 +24C +6 ⇒ 24C = 0 ⇒ C= 0$

Y(s) = $\frac{\frac{1}{4}}{s-1} + \frac{\frac{-1}{4}}{s+1} + \frac{\frac{1}{2}}{s^2+1}$

**Step 5: Apply the Inverse Laplace Transform**

Using a table of transforms, we know: $\mathcal{L}^{-1}(\frac{a}{s^2+a^2}) = sin(at), \mathcal{L}^{-1}(\frac{1}{s-a}) = e^{at}$

Combining these results, the solution to the IVP is: y(t) = $\frac{1}{4}e^t-\frac{1}{4}e^{-t}+\frac{1}{2}sin(t)$

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License and is based on MIT OpenCourseWare [18.01 Single Variable Calculus, Fall 2007].

- NPTEL-NOC IITM, Introduction to Galois Theory.
- Algebra, Second Edition, by Michael Artin.
- LibreTexts, Calculus and Calculus 3e (Apex). Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
- Field and Galois Theory, by Patrick Morandi. Springer.
- Michael Penn, and MathMajor.
- Contemporary Abstract Algebra, Joseph, A. Gallian.
- YouTube’s Andrew Misseldine: Calculus. College Algebra and Abstract Algebra.
- MIT OpenCourseWare [18.03 Differential Equations, Spring 2006], YouTube by MIT OpenCourseWare.
- Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.