Question

Given \(\mathcal{L}\\{x\\}=X, x(0)=2, x^{\prime}(0)=3\) find the Laplace transform of each of the following expressions: (a) \(x^{\prime}\) (b) \(x^{\prime \prime}\) (c) \(2 x^{\prime}-x\) (d) \(3 x^{\prime \prime}+2 x^{\prime}-x\) (e) \(-2 x^{\prime \prime}+x^{\prime}+3 x\)

Short answer

a. x'(t) b. x''(t) c. 2x'(t) - x(t) d. 3x''(t) + 2x'(t) - x(t) e. -2x''(t) + x'(t) + 3x(t) Answer: a. The Laplace transform of x'(t) is \(sX(s)-2\). b. The Laplace transform of x''(t) is \(s^2X(s)-2s-3\). c. The Laplace transform of 2x'(t) - x(t) is \(2sX(s) - 4 - X(s)\). d. The Laplace transform of 3x''(t) + 2x'(t) - x(t) is \(3s^2X(s) - 6s - 9 + 2sX(s) - 4 - X(s)\). e. The Laplace transform of -2x''(t) + x'(t) + 3x(t) is \(-2s^2X(s) + 4s + 6 + sX(s) - 2 + 3X(s)\).

Step-by-step solution

Find the Laplace Transform of x'(t) (Expression a)

We're given that \(\mathcal{L}\\{x(t)\\} = X(s)\). Applying the Laplace transform to the derivative x'(t), we have: $$ \mathcal{L}\\{x'(t)\\} = sX(s) - x(0) $$ Now, substitute the given initial condition x(0) = 2: $$ \mathcal{L}\\{x'(t)\\} = sX(s) -2 $$ So, the Laplace transform of x'(t) is \(sX(s)-2\).

Find the Laplace Transform of x''(t) (Expression b)

Applying the Laplace transform to the second derivative x''(t), we have: $$ \mathcal{L}\\{x''(t)\\} = s^2X(s) - sx(0) - x'(0) $$ Now, substitute the given initial conditions x(0) =2 and x'(0) =3: $$ \mathcal{L}\\{x''(t)\\} = s^2X(s) - 2s - 3 $$ So, the Laplace transform of x''(t) is \(s^2X(s)-2s-3\).

Find the Laplace Transform of 2x'-x (Expression c)

Now, we need to find the Laplace transform of the expression 2x'(t) - x(t): $$ \mathcal{L}\\{2x'(t) - x(t)\\} = 2\mathcal{L}\\{x'(t)\\} - \mathcal{L}\\{x(t)\\} $$ Substitute the Laplace transforms we found in steps 1 and 2, and the given Laplace transform of \(x(t)\) as \(X(s)\): $$ \mathcal{L}\\{2x'(t) - x(t)\\} = 2(sX(s)-2) - X(s) $$ So, the Laplace transform of 2x'(t) - x(t) is \(2sX(s) - 4 - X(s)\).

Find the Laplace Transform of 3x''+2x'-x (Expression d)

Now, we need to find the Laplace transform of the expression 3x''(t) + 2x'(t) - x(t): $$ \mathcal{L}\\{3x''(t) + 2x'(t) - x(t)\\} = 3\mathcal{L}\\{x''(t)\\} + 2\mathcal{L}\\{x'(t)\\} - \mathcal{L}\\{x(t)\\} $$ Substitute the Laplace transforms we found in steps 1, 2, and the given Laplace transform of x(t) as X(s): $$ \mathcal{L}\\{3x''(t) + 2x'(t) - x(t)\\} = 3(s^2X(s) - 2s - 3) + 2(sX(s)-2) - X(s) $$ So, the Laplace transform of 3x''(t) + 2x'(t) - x(t) is \(3s^2X(s) - 6s - 9 + 2sX(s) - 4 - X(s)\).

Find the Laplace Transform of -2x''+x'+3x (Expression e)

Now, we need to find the Laplace transform of the expression -2x''(t) + x'(t) + 3x(t): $$ \mathcal{L}\\{-2x''(t) + x'(t) + 3x(t)\\} = -2\mathcal{L}\\{x''(t)\\} + \mathcal{L}\\{x'(t)\\} + 3\mathcal{L}\\{x(t)\\} $$ Substitute the Laplace transforms we found in steps 1, 2, and the given Laplace transform of x(t) as X(s): $$ \mathcal{L}\\{-2x''(t) + x'(t) + 3x(t)\\} = -2(s^2X(s) - 2s - 3) + (sX(s)-2) + 3X(s) $$ So, the Laplace transform of -2x''(t) + x'(t) + 3x(t) is \(-2s^2X(s) + 4s + 6 + sX(s) - 2 + 3X(s)\).

Key concepts

Initial Conditions in Laplace Transforms

Initial conditions play a crucial role when using Laplace transforms to solve differential equations. They provide the necessary starting values for a function and its derivatives, allowing us to uniquely determine the solution later on. Often, initial conditions are represented as the value of the function and its first few derivatives at time \(t = 0\).

When we apply the Laplace transform to a differential equation, these initial conditions help us transition from the time domain to the s-domain. Using our knowledge from calculus, the Laplace transform of the first derivative \(x'(t)\) is expressed as:

• \(\mathcal{L}\{x'(t)\} = sX(s) - x(0)\)

The equation intuitively shows how the initial condition \(x(0)\) modifies the Laplace transform. For example, in our exercise, \(x(0) = 2\). Thus, when we compute \(\mathcal{L}\{x'(t)\}\), it becomes \(sX(s) - 2\), clearly indicating how the initial value is used.

Similarly, with the second derivative, \(x''(t)\), the Laplace transform is:

• \(\mathcal{L}\{x''(t)\} = s^2X(s) - sx(0) - x'(0)\)

Here, both initial values \(x(0) = 2\) and \(x'(0) = 3\) are involved, demonstrating how important these conditions are to arriving at the correct solution.

Understanding Differential Equations

Differential equations are equations that involve derivatives of a function. They are vital tools used in modeling situations where change occurs continuously, like heat transfer, population growth, or electrical circuits. Typically, these equations describe how a particular quantity's rate of change is related to the present state of the quantity.

The fundamental objective when working with differential equations is to find an expression for the unknown function. In this context, Laplace transforms provide a powerful tool allowing us to solve these equations more easily by converting them into algebraic equations in the s-domain.

In our example exercise, we dealt with expressions involving derivatives like \(x'(t)\) and \(x''(t)\). Applying Laplace transforms, we translate these derivatives into polynomial terms in \(s\), simplifying the equations. This process helps not only solve them but also handles complex initial value problems gracefully. The transformed expressions simplify handling both homogeneous and non-homogeneous equations efficiently.

S-Domain Analysis in Laplace Transforms

S-domain analysis refers to the method of examining systems and functions in the Laplace transform domain, or the s-domain. This transformation links the time domain, where functions depend on time \(t\), to the s-domain, which is a complex frequency domain. The primary advantage is how it transforms differential operations into algebraic forms.

In the s-domain, a problem that initially seems complex due to time dependencies can be tackled using algebraic methods. For instance, derivatives become simple multiplications by \(s\), and initial conditions appear as constants, which we can add or subtract. Consider our earlier transforms:

• \(\mathcal{L}\{x'(t)\} = sX(s) - x(0)\)
• \(\mathcal{L}\{x''(t)\} = s^2X(s) - sx(0) - x'(0)\)

The s-domain representation is crucial. It converts a differential equation into a format that's easier to solve. After finding the solution in the s-domain, inverse Laplace transforms allow us to convert solutions back to the time domain. This powerful technique makes it possible to not only solve for the function but also analyze system characteristics like stability and frequency response.