Problem 63 - Advanced Engineering Mathematics

Question

In Problems, use the Laplace transform to solve the given initial-value problem. $$ y^{\prime}+y=f(t), \quad y(0)=0, \text { where } f(t)=\left\\{\begin{array}{lr} 0, & 0 \leq t<1 \\ 5, & t \geq 1 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \( y(t) = 5(1-e^{-(t-1)})u(t-1) \).

Step 1: Write the Differential Equation

The initial-value problem is given as \( y'+y=f(t) \). Here, \( y(0)=0 \) is the initial condition and \( f(t) \) is a piecewise function.

Step 2: Express f(t) Using Unit Step Function

The function \( f(t) \) can be expressed using the unit step function as \( f(t) = 5u(t-1) \), where \( u(t-1) \) steps from 0 to 1 at \( t=1 \).

Step 3: Take the Laplace Transform

Take the Laplace Transform of both sides of the differential equation. The transform of \( y'(t) \) is \( sY(s) - y(0) \), and for \( y(t) \) it is \( Y(s) \). Thus, the left side becomes \( sY(s) + Y(s) = (s+1)Y(s) \). On the right side, the transform of \( f(t) = 5u(t-1) \) is \( \frac{5e^{-s}}{s} \).

Step 4: Solve for Y(s)

Solve the transformed equation \( (s+1)Y(s) = \frac{5e^{-s}}{s} \). This gives \( Y(s) = \frac{5e^{-s}}{s(s+1)} \).

Step 5: Partial Fraction Decomposition

Decompose \( \frac{5}{s(s+1)} \) into partial fractions: \( \frac{5}{s(s+1)} = \frac{A}{s} + \frac{B}{s+1} \). Solving gives \( A = 5 \) and \( B = -5 \), so \( Y(s) = 5\left(\frac{1}{s} - \frac{1}{s+1}\right)e^{-s} \).

Step 6: Inverse Laplace Transform

Find the inverse Laplace transform to get \( y(t) \). Using the properties of the Laplace transform, we have \( L^{-1}\left(\frac{1}{s}e^{-s}\right) = u(t-1) \) and \( L^{-1}\left(\frac{1}{s+1}e^{-s}\right) = e^{-(t-1)}u(t-1) \). Thus, \( y(t) = 5(u(t-1) - e^{-(t-1)}u(t-1)) \).

Step 7: Write the Final Expression for y(t)

Simplify the final expression for \( y(t) \): \( y(t) = 5(1-e^{-(t-1)})u(t-1) \).

Key Concepts

Initial Value Problems

Initial value problems (IVPs) are differential equations accompanied by specific values at the start of the equation, known as initial conditions. These conditions allow you to find particular solutions that would otherwise be part of a broader family of solutions. In the original exercise, we have an initial condition of \( y(0)=0 \). This means that when the value of \( t \) is equal to zero, \( y \) must equal zero. By imposing these specific conditions, we can solve the differential equation more accurately and get a precise solution, tailored to the scenario depicted by the initial values.
The use of Laplace transforms in solving IVPs helps to convert differential equations into an algebraic form, often making them easier to handle. Once the equation is solved in the Laplace domain, we can convert it back into the time domain to find the specific solution that fits the given initial condition.

Unit Step Function

The unit step function, often denoted as \( u(t-a) \), is an essential mathematical tool for expressing piecewise functions in terms of a single formula. It "steps" from 0 to 1 at the point \( t = a \). This function is particularly useful when dealing with time-dependent changes, such as switching on at a certain moment, which is critical in the original problem's formulation.
In the given exercise, the piecewise function \( f(t) \) is expressed using a unit step function as \( f(t) = 5u(t-1) \). This notation simplifies the representation of \( f(t) \), indicating that the function is zero for \( t < 1 \) and becomes a constant 5 when \( t \geq 1 \).

• Before \( t = 1 \), \( f(t) = 0 \).
• At \( t = 1 \) and onwards, \( f(t) = 5 \).

This calculation allows us to apply Laplace transforms efficiently, as the unit step function has a straightforward Laplace transform which we leverage in further steps of solving the problem.

Inverse Laplace Transform

The inverse Laplace transform is the process by which we convert an algebraic expression in the frequency domain back into a function in the time domain. After solving the differential equation using Laplace transforms, the solution is typically presented as a function of \( s \), denoted as \( Y(s) \). To find \( y(t) \), we apply the inverse transform, translating the solution back to the original variable, \( t \).
In the context of the solution, after finding \( Y(s) \), we need to employ the inverse Laplace transform to arrive at \( y(t) \). This is where our understanding of standard transform pairs becomes crucial. For instance, knowing that \( L^{-1}(\frac{1}{s}e^{-as}) = u(t-a) \) helps us find that \( L^{-1}(\frac{1}{s+1}e^{-as}) \) corresponds to \( e^{-(t-a)}u(t-a) \), allowing us to solve specific time-delay components.
The resulting expression \( y(t) = 5(1-e^{-(t-1)})u(t-1) \) effectively applies these inverse transformations, giving the final solution in terms of \( t \).

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to break down complex rational expressions into simpler fractions that are easier to work with, especially when taking Laplace transforms or their inverses. It is a crucial step in finding the inverse Laplace transform of functions that result from solving differential equations.
For the exercise, after finding \( Y(s) = \frac{5e^{-s}}{s(s+1)} \), we decompose \( \frac{5}{s(s+1)} \) into simpler fractions: \( \frac{A}{s} + \frac{B}{s+1} \). This step involves determining the coefficients \( A \) and \( B \) by solving the equation \( \frac{5}{s(s+1)} = \frac{A}{s} + \frac{B}{s+1} \). Upon solving, we find \( A = 5 \) and \( B = -5 \), resulting in:\[Y(s) = 5\left(\frac{1}{s} - \frac{1}{s+1}\right)e^{-s}\]
This decomposition simplifies the inversion process, as it allows us to use well-known inverse Laplace transforms for each separate term, which are key to constructing the time-domain solution.