Question

Use the Laplace transforms to solve each of the initial-value. \(y^{\prime}-y=e^{3 t}\) \(y(0)=2\)

Short answer

The solution for the initial value problem is: \(y(t) = 2e^t+\frac{1}{2}e^{3t}-\frac{1}{2}e^t\)

Step-by-step solution

Apply the Laplace transform

First, let's take the Laplace transform of the given differential equation: \(y'(t) - y(t) = e^{3t}\) Taking the Laplace transform of both sides, we get: \(\mathcal{L}\{y'(t) - y(t)\} = \mathcal{L}\{e^{3t}\}\) Applying the Laplace transform rules on both sides, we have: \(sY(s) - y(0) - Y(s) = \frac{1}{s-3}\)

Solve for Y(s)

Now, let's solve this algebraic equation for Y(s): \(sY(s) - Y(s) = y(0) + \frac{1}{s-3}\) \(Y(s) = \frac{y(0)}{s-1} + \frac{1}{(s-3)(s-1)}\) Given that \(y(0) = 2\), we can substitute this value into the equation above to find Y(s): \(Y(s) = \frac{2}{s-1} + \frac{1}{(s-3)(s-1)}\)

Apply the inverse Laplace transform

Now, we apply the inverse Laplace transform to Y(s) to find y(t): \(y(t) = \mathcal{L}^{-1}\{Y(s)\}\) \(y(t) = \mathcal{L}^{-1}\{\frac{2}{s-1} + \frac{1}{(s-3)(s-1)}\}\) Using the linearity property of inverse Laplace transforms, we have: \(y(t) = 2\mathcal{L}^{-1}\{\frac{1}{s-1}\} + \mathcal{L}^{-1}\{\frac{1}{(s-3)(s-1)}\}\) The inverse Laplace transforms of \(\frac{1}{s-1}\) and \(\frac{1}{(s-3)(s-1)}\) are known: \(\mathcal{L}^{-1}\{\frac{1}{s-1}\} = e^{t}\) \(\mathcal{L}^{-1}\{\frac{1}{(s-3)(s-1)}\} = \frac{1}{2}e^{3t} - \frac{1}{2}e^t\) Substitute these into the equation for y(t): \(y(t) = 2e^t + \left(\frac{1}{2}e^{3t} - \frac{1}{2}e^t\right)\)

Check the initial condition

Finally, let's ensure that the solution we found satisfies the initial condition \(y(0) = 2\). Plug in \(t=0\) into the solution: \(y(0) = 2e^0 + \left(\frac{1}{2}e^{3 \cdot 0} - \frac{1}{2}e^0\right)\) \(y(0) = 2 + \left(\frac{1}{2} - \frac{1}{2}\right)\) As you can see, the initial condition is satisfied: \(y(0) = 2\) The solution for the initial value problem is: \(y(t) = 2e^t+\frac{1}{2}e^{3t}-\frac{1}{2}e^t\)

Key concepts

Ordinary Differential Equations

Ordinary differential equations (ODEs) form a subset of differential equations containing one or more unknown functions of a single variable and their derivatives. In our exercise, the problem is defined by the equation \(y' - y = e^{3t}\). A key characteristic of an ODE is that it involves the derivative of an unknown function and perhaps the function itself. These problems frequently describe physical phenomena, such as motion, growth, or decay.
ODEs can vary in order, complexity, and linearity. The order is determined by the highest derivative present in the equation. In our exercise, the differential equation is of the first order since it involves the first derivative \(y'\). Linear equations are where both the function and its derivatives appear linearly, as shown in \(y' - y\). Such equations are often more straightforward to solve using various methods, including Laplace transforms.
Understanding the form and characteristics of an ODE is essential for effectively approaching and solving these problems.

Initial Value Problem

An initial value problem is a type of differential equation that provides additional conditions known as initial conditions. These initial conditions specify the value of the unknown function at a particular point, which is crucial for determining a unique solution to the differential equation.
In our example, the initial condition provided is \(y(0) = 2\). This means that at the time \(t=0\), the value of \(y\) is 2. Initial conditions help to anchor the solution of the differential equation in the realm of real-world applications, ensuring the solution is specific to a given scenario.
To solve an initial value problem, computation generally involves:

• Writing down the differential equation and initial conditions.
• Using an appropriate solving method, such as the Laplace transform, to handle the equation.
• Applying the initial conditions to conclude the specific solution.

Successfully applying the initial conditions ensures consistency and applicability in problems across mathematics, engineering, and the sciences.

Inverse Laplace Transform

The inverse Laplace transform is a critical tool for finding solutions to differential equations converted into algebraic equations by the Laplace transform. Once an expression is transformed using the Laplace method, the aim is to apply the inverse transform to arrive back at the original functional form.
In our solution process, we reached an expression for \(Y(s)\) that needed reverting back to \(y(t)\). Employing the inverse Laplace transform involves reversing this process, effectively transforming each part of \(Y(s)\) back into time-domain components.

• The inverse transform of \(\frac{1}{s-1}\) corresponds to \(e^t\).
• The inverse transform of \(\frac{1}{(s-3)(s-1)}\) is more complex, resulting in the expression \(\frac{1}{2}e^{3t}-\frac{1}{2}e^t\).

Mastering the inverse Laplace transform technique allows for resolving these kinds of differential equations more efficiently, translating complex algebraic results back into meaningful time-based functions.

Laplace Transform Rules

Laplace transform rules simplify the process of solving differential equations by converting them into algebraic equations. Certain key rules facilitate this transformation, making the Laplace method a powerful analytical tool.
The primary rule used in our problem is related to the transformation of derivatives. For a first derivative, \(y'(t)\), the Laplace transform is \(sY(s) - y(0)\). This allows a differential equation to become algebraically solvable when coupled with other transformations.
Additionally, using transform tables or known transform rules helps to convert complex functions into simpler algebraic forms during analysis.

• Linearity: \(\mathcal{L}\{a f(t) + b g(t)\} = a \mathcal{L}\{f(t)\} + b \mathcal{L}\{g(t)\}\).
• Exponential shift: \(\mathcal{L}\{e^{at}f(t)\} = F(s-a)\).

Understanding and applying these rules effectively is crucial for handling complex equations, where direct solving methods may not be feasible. The rules underpin the reliable conversion process between time and frequency domains in mathematical analysis.