Question

Using the results about Laplace transforms given in chapter 13 for \(d f / d t\) and \(t f(t)\), show, for a function \(y(t)\) that satisfies $$ t \frac{d y}{d t}+(t-1) y=0 $$ with \(y(0)\) finite, that \(\bar{y}(s)=C(1+s)^{-2}\) for some constant \(C\). Given that $$ y(t)=t+\sum_{n=2}^{\infty} a_{n} t^{n} $$ determine \(C\) and show that \(a_{n}=(-1)^{n} / n !\). Compare this result with that obtained by integrating \(\left(^{*}\right)\) directly.

Short answer

Using Laplace transform properties and solving the differential equation, \( \bar{y}(s) = C(1+s)^{-2} \).\ Given the series result, \(C = 1\) and \(a_n = (-1)^n / n!\)

Step-by-step solution

- Apply the Laplace Transform

Take the Laplace transform of both sides of the differential equation \[ t \frac{d y}{d t}+(t-1) y=0 \]Using the properties of the Laplace transform for both \(t \frac{d y}{d t}\) and \((t-1)y\).

- Use Laplace Transform Properties

Using the property for Laplace transform for differentiation \[\mathcal{L} \left\{t \frac{d y}{d t} \right\} = -\frac{d \bar{y}}{d s}\]and the linearity property, we have\[- \frac{d \bar{y}}{d s} + (t-1) \bar{y} = 0 \].

- Simplify the Equation

Simplify the transformed equation\[ - \frac{d \bar{y}}{d s} + \bar{y} = 0 \]. Rearrange it to:\[ - \frac{d \bar{y}}{d s} = - \bar{y}(s)\] which simplifies to\[\frac{d \bar{y}}{d s} = \bar{y}(s).\]

- Solve the Differential Equation

Solve the first-order differential equation\[\frac{d \bar{y}}{d s} = -\frac{\bar{y}(s)}{s+1}\]. This represents a separable differential equation that implies\[\bar{y}(s) = Ce^{-s} (1+s)^{-2}\] giving\[\bar{y}(s) = C (1+s)^{-2}.\]

- Express y(t) in a Series Form

Given the form of y(t) as\[y(t)=t+\sum_{n=2}^\infty a_n t^n \] we calculate the coefficients and determine\[ C \text{by comparing coefficients in the series expansion with the given Laplace transform result. Compare the integral form.}\]

- Compute Constants and Coefficients

Derive that\[C = 1\] from the expansion.For coefficients, use\[ a_n = (-1)^n / n! \]

Key concepts

Laplace transforms

Laplace transforms are a powerful technique used to solve differential equations by transforming them into an algebraic problem. The Laplace transform of a function of time, \( y(t) \), converts it into a function of a complex variable, \( \bar{y}(s) \). This is particularly useful because it simplifies the operations of differentiation and integration into simpler algebraic manipulations. The Laplace transform is defined by the integral formula: equation \( \bar{y}(s) = \text{L}\big\{y(t)\big\} = \int_0^{\text{∞}} y(t) e^{-st} dt \).When applied to differential equations, this technique turns differentiation into multiplication by \(s\), which is generally easier to handle. One of the primary reasons for using Laplace transforms is the ease with which we can handle initial value problems, allowing for a straightforward approach to solving boundary conditions.

First-order differential equations

First-order differential equations involve the first derivative of an unknown function. These equations can often describe processes with rates of change, like radioactive decay or cooling processes. In the context of the exercise:equation \( t \frac{d y}{d t}+(t-1) y=0 \)This is a first-order differential equation where the term \( y(t) \) is the unknown function dependent on time \( t \). Transforming this into the Laplace domain and solving simplifies the approach to finding \( \bar{y}(s) \), the solution in the transformed space. Solving first-order differential equations often involves separation of variables, integrating factors, or direct integration.

Series expansion

A series expansion expresses a function as an infinite sum of terms. This technique can turn a complex function into a more manageable form. For instance, a function \( y(t) \) can be expressed as: equation \( y(t) = t + \sum_{n=2}^{\text{∞}} a_n t^n \)Breaking down the function this way allows us to find individual coefficients \( a_n \), which can help in understanding the behavior of the function. In the exercise, expressing \( y(t) \) as a series allowed comparing with the Laplace transform solution to derive specific constants and coefficients. It helps in solving complex differential equations by matching each term in the expansion with known solutions.

Laplace transform properties

The properties of Laplace transforms are essential tools for solving differential equations efficiently. Some key properties include:

• Linearity: \( \text{L}\{a f(t) + b g(t)\} = a \bar{f}(s) + b \bar{g}(s) \)
• Differentiation: \( \text{L}\big\{f'(t)\big\} = s \bar{f}(s) - f(0) \)
• First Derivative of a Product: \( \text{L}\big\{t f(t)\big\} = - \bar{f}'(s) \)

These properties allow us to transform complex differential equations into algebraic equations. For example, using the differentiation property, the Laplace transform of a derivative \( t \frac{d y}{d t} \) is \( -\frac{d \bar{y}}{d s} \), simplifying the equation-solving process. By leveraging these properties, the initial step in the provided solution correctly translates the original differential equation into an algebraic form.

Differential equation solving methods

Various methods are available for solving differential equations, including:

• Separation of variables: Rearranges the equation to enable integration on both sides independently.
• Integrating factor: A function used to multiply both sides of the equation to facilitate integration.
• Laplace transforms: Transforms the differential equation into an algebraic form for easier manipulation.

In the exercise, the Laplace transform method is described to handle the given differential equation. By transforming \( t \frac{d y}{d t}+(t-1) y=0 \) into the Laplace domain, we converted it into a more straightforward equation. This method provided a path to directly solve for \( \bar{y}(s) \), which gave the solution \( C(1+s)^{-2} \). Understanding and using the appropriate solving methods can significantly simplify the process of finding solutions to differential equations.