Question

Find a formula involving integrals for a particular solution of the differential equation $$ y^{\prime \prime \prime}-3 y^{\prime \prime}+3 y^{\prime}-y=g(t) $$ If \(g(t)=t^{-2} e^{t},\) determine \(Y(t)\)

Short answer

The purpose of using the convolution theorem in this exercise is to rewrite the Laplace transform of the product of two functions as the integral of products of their inverse Laplace transforms, which facilitates finding the particular solution Y(t) for the given differential equation.

Step-by-step solution

Rewrite the given differential equation

We are given the differential equation: $$ y^{\prime \prime \prime}-3 y^{\prime \prime}+3 y^{\prime}-y=g(t) $$ where \(g(t)=t^{-2} e^{t}\)

Take Laplace transform of the equation

Taking the Laplace transform on both sides of the equation, we get: $$ \Lap[ y^{\prime \prime \prime}] - 3\Lap[y^{\prime \prime}] + 3\Lap[y^{\prime}] - \Lap[y] = \Lap[g(t)] $$ Using the Laplace transform properties, we can rewrite the equation as: $$ s^3Y(s) - 3 s^2 Y(s) + 3s Y(s) - Y(s) = \lap[ t^{-2} e^{t}] $$

Find the Laplace transform of a particular solution using convolution theorem

Using the convolution theorem, we can rewrite the equation as: $$ ( s^3 - 3 s^2 + 3s - 1) Y(s) = \frac{ e^s \Gamma(0,s) } {s^2} $$ Now, we need to find the inverse Laplace transform to get \(Y(t)\): $$ y(t) = L^{-1}\left( \frac{ e^s \Gamma(0,s) } {s^2(s^3 - 3 s^2 + 3s - 1)}\right) $$

Use convolution theorem and inverse Laplace transforms to determine \(Y(t)\)

Using convolution theorem, we have: $$ y(t) = \int_0^t L^{-1}\left( \frac{ e^s \Gamma(0,s) } {s^2} \right) L^{-1}(s^3 - 3 s^2 + 3 s - 1)ds $$ For convenience, let \(u(s) = L^{-1}\left( \frac{ e^s \Gamma(0,s) } {s^2} \right)\) and \(v(s) = L^{-1}(s^3 - 3 s^2 + 3 s - 1)\). Then: $$ y(t) = \int_0^t u(s) \cdot v(t-s) ds $$ Now, we need to find the inverse Laplace transforms of \(u(s)\) and \(v(s)\) and then calculate the integral. Note that we may need to use partial fraction decomposition or other Laplace transform properties and techniques to find the inverse Laplace transforms. After finding the inverse Laplace transforms and calculating the integral, we will have our particular solution \(Y(t)\).

Key concepts

Laplace Transform

The Laplace Transform is a powerful tool used in solving differential equations, particularly those with boundary value problems. It's a method for transforming a complex differential equation into a simpler algebraic equation. The Laplace Transform of a function \(f(t)\), where \(t \/ geq 0\), is defined as:\[\Lap[f(t)] = \int_0^\infty e^{-st}f(t) \,dt\]This transformation shifts the problem from the time domain into the s-domain (complex frequency domain), making it easier to solve for functions that have derivatives. One key property that makes the Laplace Transform so useful is its ability to convert differential equations into algebraic ones.

• The transform of a derivative \(f'(t)\) is \(s \Lap[f(t)] - f(0)\).
• Similarly, \(\Lap[f''(t)] = s^2 \Lap[f(t)] - sf(0) - f'(0)\).

These properties transform the given differential equation into an algebraic equation when applied. Solving this algebraic equation using algebraic methods is often more straightforward.

Convolution Theorem

The Convolution Theorem is an essential concept in the realm of Laplace Transforms and differential equations. This theorem provides a way to handle the inverse transformation of products of Laplace Transforms. It states that:

• If \( L[f(t)] = F(s) \) and \( L[g(t)] = G(s) \), then the Laplace Transform of their convolution is the product of their transforms: \( L[f * g](s) = F(s)G(s) \).
• The convolution itself is defined as \((f * g)(t) = \int_0^t f(\tau) g(t - \tau) d\tau\).

In practice, this allows the inverse Laplace transform of a product of functions to be expressed as a convolution integral of their respective inverse transforms. When solving the differential equation in the problem, using the convolution theorem simplifies finding the particular solution by circumventing direct inverse Laplace transforms of complex products.

Inverse Laplace Transform

The Inverse Laplace Transform is the process of converting a function in the s-domain back into the time domain. This is typically done using tables of known transforms or mathematical techniques such as partial fraction decomposition. For any function \(F(s)\), its inverse transform is:\[y(t) = \Lap^{-1}[F(s)]\]To determine \(y(t)\), we often match \(F(s)\) to known Laplace Transforms or manipulate it into a form that reveals its inverse. In the case of finding a particular solution, as in our problem, it's often necessary to break down or simplify \(F(s)\) using techniques like:

• Partial fraction decomposition, which separates complex fractions into simpler terms.
• Convolution Theorem, which links products in the s-domain to convolutions in the time domain.

By applying these techniques effectively, we translate the algebraic solution back into the differential function that satisfies the original equation.

Particular Solution

In solving differential equations, particularly with non-homogeneous terms, the goal is often to find a particular solution that satisfies the entire equation. The particular solution is a specific solution to a differential equation that includes the non-homogeneous part, represented by \(g(t)\) in our problem.

• The particular solution combines with the homogeneous solution to give the general solution of the equation.
• It usually requires specific techniques such as undetermined coefficients or variation of parameters.

In Laplace Transform methods, we find the particular solution by transforming the differential equation, solving the resultant algebraic equation, and inversely transforming the solution. Often, as in the given exercise, convolution and inverse transformation processes are used in tandem to form the particular solution. This process results in a real-world applicable solution that fits the given initial conditions and non-homogeneous parts of the differential equation.