Solving For Y(s) In Laplace Transform Of Initial Value Problem

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#Title: Solving for Y(s) Using Laplace Transforms in an Initial Value Problem

This article delves into the process of solving for Y(s), which represents the Laplace transform of the solution y(t), in a given initial value problem. We'll break down the steps involved, focusing on applying Laplace transforms to differential equations and utilizing initial conditions to arrive at the solution in the s-domain. This is a crucial technique in engineering and physics for analyzing dynamic systems.

Problem Statement

Let's consider the following initial value problem:

  • y''(t) - 9y'(t) + 18y(t) = 2te^(3t)
  • y(0) = 6
  • y'(0) = -5

Our goal is to find Y(s), the Laplace transform of y(t). To achieve this, we will systematically apply Laplace transforms to each term in the differential equation, incorporate the initial conditions, and then solve for Y(s). This process transforms the differential equation into an algebraic equation in the s-domain, making it easier to solve.

Applying Laplace Transforms

Laplace Transform of Derivatives

The Laplace transform has a powerful property when dealing with derivatives. Specifically:

  • L{y'(t)} = sY(s) - y(0)
  • L{y''(t)} = s^2Y(s) - sy(0) - y'(0)

These properties are essential for transforming the differential equation into the s-domain. They allow us to replace derivatives with algebraic expressions involving Y(s) and the initial conditions. Understanding these transformations is key to solving differential equations using Laplace transforms.

Laplace Transform of the Forcing Function

The forcing function in our problem is 2te^(3t). We need to find its Laplace transform. Using the Laplace transform table (as referenced in the prompt) or applying the definition of the Laplace transform, we find:

  • L{2te^(3t)} = 2/(s-3)^2

This result is crucial because it represents the input or driving force of the system in the s-domain. It highlights how the exponential function with a time-dependent term translates into a rational function in the Laplace domain, which is easier to manipulate algebraically.

Transforming the Entire Equation

Now, we apply the Laplace transform to the entire differential equation:

L{y''(t) - 9y'(t) + 18y(t)} = L{2te^(3t)}

Using the linearity property of the Laplace transform, we can write:

L{y''(t)} - 9L{y'(t)} + 18L{y(t)} = L{2te^(3t)}

Substituting the Laplace transforms of the derivatives and the forcing function, we get:

[s^2Y(s) - sy(0) - y'(0)] - 9[sY(s) - y(0)] + 18Y(s) = 2/(s-3)^2

Incorporating Initial Conditions

We are given the initial conditions y(0) = 6 and y'(0) = -5. We substitute these values into the equation obtained in the previous step:

[s^2Y(s) - 6s - (-5)] - 9[sY(s) - 6] + 18Y(s) = 2/(s-3)^2

Simplifying the equation, we have:

s^2Y(s) - 6s + 5 - 9sY(s) + 54 + 18Y(s) = 2/(s-3)^2

Solving for Y(s)

Now, we need to isolate Y(s). First, group the terms containing Y(s):

(s^2 - 9s + 18)Y(s) - 6s + 59 = 2/(s-3)^2

Next, move the non-Y(s) terms to the right side of the equation:

(s^2 - 9s + 18)Y(s) = 2/(s-3)^2 + 6s - 59

Now, factor the quadratic term and find a common denominator on the right side:

(s - 3)(s - 6)Y(s) = [2 + (6s - 59)(s - 3)^2] / (s - 3)^2

Finally, divide both sides by (s - 3)(s - 6) to solve for Y(s):

Y(s) = [2 + (6s - 59)(s - 3)^2] / [(s - 3)^2 (s - 3)(s - 6)]

Simplifying Y(s)

To simplify Y(s), we expand the numerator:

Y(s) = [2 + (6s - 59)(s^2 - 6s + 9)] / [(s - 3)^3 (s - 6)]

Expanding further:

Y(s) = [2 + 6s^3 - 36s^2 + 54s - 59s^2 + 354s - 531] / [(s - 3)^3 (s - 6)]

Combining like terms:

Y(s) = (6s^3 - 95s^2 + 408s - 529) / [(s - 3)^3 (s - 6)]

This is the expression for Y(s), the Laplace transform of the solution y(t) to the given initial value problem. This expression provides a representation of the solution in the s-domain, which can be further analyzed or inverted back into the time domain to find y(t). The process of simplifying and manipulating algebraic expressions in the s-domain is a critical step in solving differential equations using Laplace transforms.

Partial Fraction Decomposition (Optional)

To find y(t), we would typically perform partial fraction decomposition on Y(s) and then use the inverse Laplace transform. This step is optional for this problem since we were only asked to find Y(s). However, it's important to understand that partial fraction decomposition is often necessary to convert Y(s) back into a time-domain function.

Key Concepts

Laplace Transform

The Laplace transform is a mathematical tool that transforms a function of time, t, into a function of a complex variable, s. This transformation is particularly useful for solving linear differential equations because it converts them into algebraic equations, which are often easier to solve. The Laplace transform is defined as:

L{f(t)} = F(s) = ∫[0 to ∞] e^(-st) f(t) dt

Where:

  • f(t) is the function of time.
  • F(s) is the Laplace transform of f(t).
  • s is a complex variable.

Properties of Laplace Transforms

The effectiveness of Laplace transforms in solving differential equations stems from several key properties. These properties allow us to manipulate and simplify equations in the s-domain, making them easier to solve.

Linearity

The Laplace transform is a linear operator, meaning that it satisfies the following properties:

  • L{af(t)} = aL{f(t)}, where a is a constant.
  • L{f(t) + g(t)} = L{f(t)} + L{g(t)}

These properties allow us to transform sums and scalar multiples of functions separately and then combine the results. This is particularly useful when dealing with complex equations that involve multiple terms.

Transform of Derivatives

One of the most significant properties of the Laplace transform is its ability to transform derivatives into algebraic expressions. This property is crucial for converting differential equations into algebraic equations. The Laplace transforms of the first and second derivatives are given by:

  • L{y'(t)} = sY(s) - y(0)
  • L{y''(t)} = s^2Y(s) - sy(0) - y'(0)

Where:

  • y'(t) and y''(t) are the first and second derivatives of y(t), respectively.
  • Y(s) is the Laplace transform of y(t).
  • y(0) and y'(0) are the initial conditions of y(t) and y'(t), respectively.

These formulas show that the Laplace transform converts differentiation in the time domain into multiplication by s in the s-domain, along with terms involving initial conditions. This is a powerful simplification that makes solving differential equations much easier.

Transform of Integrals

The Laplace transform can also simplify integrals. The transform of the integral of a function is given by:

L{∫[0 to t] f(τ) dτ} = (1/s)F(s)

This property states that integrating a function in the time domain corresponds to dividing its Laplace transform by s in the s-domain. This is useful for solving integral equations and integro-differential equations.

Time Shifting

The time-shifting property, also known as the translation property, states that if L{f(t)} = F(s), then:

L{f(t - a)u(t - a)} = e^(-as)F(s)

Where:

  • a is a constant.
  • u(t - a) is the unit step function, which is 0 for t < a and 1 for t ≄ a.

This property is useful for dealing with time delays in systems. It shows that shifting a function in time corresponds to multiplying its Laplace transform by an exponential term in the s-domain.

Frequency Shifting

The frequency-shifting property states that if L{f(t)} = F(s), then:

L{e^(at)f(t)} = F(s - a)

Where a is a constant. This property is useful for dealing with functions that are multiplied by an exponential term. It shows that multiplying a function by an exponential in the time domain corresponds to shifting its Laplace transform in the s-domain.

Differentiation in the s-Domain

The property of differentiation in the s-domain states that if L{f(t)} = F(s), then:

L{tf(t)} = -d/ds F(s)

This property is useful for finding the Laplace transforms of functions that are multiplied by t. It shows that multiplying a function by t in the time domain corresponds to differentiating its Laplace transform with respect to s and multiplying by -1.

Convolution

The convolution theorem is a powerful property that relates the Laplace transform of the convolution of two functions to the product of their Laplace transforms. The convolution of two functions f(t) and g(t) is defined as:

(f * g)(t) = ∫[0 to t] f(τ)g(t - τ) dτ

The convolution theorem states that:

L{(f * g)(t)} = F(s)G(s)

Where F(s) and G(s) are the Laplace transforms of f(t) and g(t), respectively. This property is particularly useful for solving integral equations and analyzing systems with complex inputs.

Initial Value Problem

An initial value problem (IVP) is a differential equation along with a set of initial conditions. The initial conditions specify the value of the function and its derivatives at a particular point (usually t = 0). These conditions are crucial for finding a unique solution to the differential equation.

Solving IVPs using Laplace Transforms

The Laplace transform method is particularly well-suited for solving linear, constant-coefficient IVPs. The general procedure involves the following steps:

  1. Transform the differential equation: Apply the Laplace transform to both sides of the differential equation. Use the properties of Laplace transforms to convert derivatives into algebraic expressions.
  2. Incorporate initial conditions: Substitute the given initial conditions into the transformed equation. This will result in an algebraic equation in terms of Y(s).
  3. Solve for Y(s): Solve the algebraic equation for Y(s). This gives the Laplace transform of the solution.
  4. Inverse transform to find y(t): Apply the inverse Laplace transform to Y(s) to obtain the solution y(t) in the time domain. This step often involves using partial fraction decomposition and looking up inverse transforms in a table.

Conclusion

In this article, we have demonstrated how to solve for Y(s), the Laplace transform of the solution to an initial value problem. We applied Laplace transforms to the differential equation, incorporated initial conditions, and solved for Y(s) in the s-domain. This process showcases the power of Laplace transforms in simplifying differential equations and providing a pathway to finding solutions. While we stopped at finding Y(s), the next step would typically involve partial fraction decomposition and inverse Laplace transformation to obtain the solution y(t) in the time domain. Understanding these concepts and techniques is essential for engineers and scientists dealing with dynamic systems and control problems. The Laplace transform is a fundamental tool in many areas of engineering, including electrical engineering, mechanical engineering, and control systems. Mastering this technique provides a powerful means for analyzing and designing dynamic systems.