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
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., $\frac{dy}{dx} = 3x +5y$):
Definition. A first-order linear ordinary differential equation is an ordinary differential equation (ODE) involving an unknown function y(x), its first derivative y′, and functions of the independent variable x, which can be written in the general form:: a(x)y' + b(x)y = c(x) where:
These equations are termed “linear” because the unknown function y and its derivative y’ appear to the first power and are not multiplied together or composed in any nonlinear way.
If the function c(x)=0 for all x in the interval of interest, the equation simplifies to: a(x)y’ + b(x)y = 0. Such an equation is called a homogeneous linear differential equation.
The Existence and Uniqueness Theorem provides crucial insight into the behavior of solutions to first-order differential equations ODEs. It states that if:
Then, the differential equation y' = f(x, y) has a unique solution to the initial value problem through the point (x0, y0), meaning that it satisfies the initial condition y(x0) = y0.
This theorem ensures that under these conditions, the solution exists and is unique near x = x0.
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) | $\mathcal{L}(u(t)) = \frac{1}{s}, s > 0$ |
$e^{at}$ | $\frac{1}{s - a}, s > a$ |
$e^{(a + bi)t}$ | $\mathcal{L}(e^{(a + bi)t}) = \frac{1}{s - (a + bi)}, s > a$ |
$\cos(\omega t)$ | $\mathcal{L}(\cos(\omega t)) = \frac{s}{s^2 + \omega^2}, s > 0$ |
$\sin(\omega t)$ | $\mathcal{L}(\sin(\omega t)) = \frac{\omega}{s^2 + \omega^2}, s > 0$ |
$t^n$ | $\mathcal{L}(t^n) = \frac{n!}{s^{n+1}}, s > 0$ |
$u(t-a)$ | $\mathcal{L}(u(t-a)) = \frac{e^{-as}}{s}, s > 0$ |
$\delta(t-a)$ | $\mathcal{L}(\delta(t-a)) = e^{-as}, a \geq 0$ |
$\frac{1}{t}$ | $\mathcal{L}\left(\frac{1}{t}\right) = \text{not defined}$ |
$e^{-bt} \cos(\omega t)$ | $\mathcal{L}(e^{-bt} \cos(\omega t)) = \frac{s + b}{(s + b)^2 + \omega^2}, s > -b$ |
$e^{-bt} \sin(\omega t)$ | $\mathcal{L}(e^{-bt} \sin(\omega t)) = \frac{\omega}{(s + b)^2 + \omega^2}, s > -b$ |
$e^{at}f(t)$ | $\mathcal{L}(e^{at}f(t)) = F(s-a)$ |
The last entry illustrates the Exponential Shift Theorem, indicating that multiplying a function by an exponential term eat shifts its Laplace Transform.
Additionally, the Laplace Transforms of derivatives are given by: $\mathcal{L}(f’(t)) = sF(s)-f(0), \mathcal{L}(f’’(t)) = s^2F(s)-sf(0)-f’(0)$
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 from the time domain into algebraic equations in the complex frequency domain, we simplify the process of finding solutions to these equations. This method is particularly useful because it converts differential equations, which involve derivatives, into algebraic equations, which are generally easier to solve.
Consider the following second-order linear ODE with constant coefficients: y’’ + Ay’ + By = h(t) where:
Our goal is to solve this equation for y(t) using the Laplace Transform. The key idea is that the Laplace Transform 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.
These formulas are derived using integration by parts and incorporate 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: $\mathcal{L}(y’’(t)+Ay’(t)+By(t)) = \mathcal{L}(h(t)) ↭ [\text{Using the Laplace Transform properties:}]\mathcal{L}(y’’(t)) + A\mathcal{L}(y’(t)) + B\mathcal{L}(y(t)) = \mathcal{L}(h(t)) ↭ [\text{Substituting the Laplace Transforms:}] (s^2Y(s) -sy_0 -y’_0) + A(sY(s) -y_0) + BY(s) = H(s)$ 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:
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:
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 and writing the general solution to the homogeneous differential equation, determining a particular solution to the ODE, and finally 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, $y’’ -y =e^{-t} ↭ (s^2Y(s) -s·Y(0) -Y’(0)) -Y(s) = \frac{1}{s+1} \leadsto (s^2Y(s) -s·1 -0) -Y(s) = \frac{1}{s+1}↭ s^2Y(s)-s -Y(s) = \frac{1}{s+1} ⇒ (s^2-1)Y = \frac{1}{s+1}+s = \frac{1 + s^2+s}{s+1}$
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: s2 +s +1 = A(s−1)+B(s+1)(s-1)+C(s+1)^2 ⇒ [Expanding terms:] s2 +s + 1 = A(s−1)+B(s2-1)+C(s2 +2s +1) =[Group terms] = (B+C)s2 +(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: (Use standard inverse Laplace Transforms $\mathcal{L}^{-1}(\frac{1}{(s-a)^2}) = te^{at}, \mathcal{L}^{-1}(\frac{1}{(s-a)}) = e^{at}$) $\frac{\frac{-1}{2}}{(s+1)^2} + \frac{\frac{1}{4}}{s+1} + \frac{\frac{3}{4}}{s-1} \leadsto \mathcal{L}^{-1}~ \frac{-1}{2}te^{-t} + \frac{1}{4}e^{-t} + \frac{3}{4}e^t$, 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
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: (s2Y(s) −sY(0) −Y′(0)) -3(sY(s)-y(0)) +2Y(s)=0 ⇒[Substitute the initial conditions] (s2Y(s) -s·1 -0) + -3(sY(s)-1) + 2Y(s) = 0 ⇒ s2Y(s) -s -3sY(s) + 3 + 2Y(s) = 0 ⇒[Collect like terms] (s2-3s +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, hence A = 2, B = -1.
$Y(s) = \frac{2}{s-1}+\frac{-1}{s-2}$
Step 4: Apply the Inverse Laplace Transform
Use standard transform, $\mathcal{L}^{-1}(\frac{1}{(s-a)}) = e^{at}$. Therefore, $y(t)= 2e^{t}-e^{2t}$
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: (s2Y(s) −sY(0) −Y′(0)) +4Y(s)=0 ⇒[Substitute the initial conditions] (s2Y(s) -s·1 -0) + 4Y(s) = 0 ⇒ (s2 +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: (s4Y(s) −s3Y(0) −s2Y′(0) -sY’’(0) -Y’’’(0)) - Y(s)=0 ⇒[Substitute the initial conditions] (s4Y(s) -s2·1) -Y(s) = 0 ⇒ s4Y(s) -s2 -Y(s) = 0.
Step 3: Solve for Y(s): $Y(s)=\frac{s^2}{s^4-1}$
Step 4. Partial Fraction Decomposition: $Y(s)=\frac{s^2}{s^4-1} = [\text{Factor the denominator}] \frac{s^2}{(s^2-1)(s^2+1)} = \frac{s^2}{(s-1)(s+1)(s^2+1)} = \frac{A}{s-1} + \frac{B}{s+1} + \frac{Cs + D}{s^2+1}$
$s^2 = A(s+1)(s^2+1) + B(s-1)(s^2+1) + (Cs+D)(s-1)(s+1)$
Evaluate at s = -1 ⇒ 1 = B(-4)⇒ B = $\frac{-1}{4}$. Evaluate at s = 1 ⇒1 = A(4) ⇒ A = $\frac{1}{4}$. Evaluate at s = 0 ⇒ $0 = \frac{1}{4}+\frac{1}{4}-D ⇒ D = \frac{1}{2}$. Evaluate at 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) = $\frac{1}{4}e^t-\frac{1}{4}e^{-t}+\frac{1}{2}sin(t)$