Chapter 15: Problem 14

By using Laplace transforms, solve the following differential equations subject to the given initial conditions. $$ y^{\prime \prime}-4 y^{\prime}=-4 t e^{2 t}, \quad y_{0}=0, \quad y_{0}^{\prime}=1 $$

Short Answer

Expert verified

The solution is obtained by taking the Laplace transform, applying initial conditions, simplifying, and then performing the inverse Laplace transform to find $ y(t)$.

Step by step solution

- Take the Laplace Transform of Both Sides of the Differential Equation

Apply the Laplace transform to each term in the given differential equation. The Laplace transform of the second derivative is $\mathcal{L}\{y''(t)\} = s^2 Y(s) - sy(0) - y'(0)$, and for the first derivative it is $\mathcal{L}\{y'(t)\} = sY(s) - y(0)$. Applying these, we get: $\text{Laplace}(y'' - 4y') = \text{Laplace}(-4te^{2t})$.

- Use Initial Conditions

Substitute the initial conditions $ y(0)=0$ and $ y'(0)=1$ into the transformed equation. This gives us: \[ s^2 Y(s) - s \times 0 - 1 - 4(sY(s) - 0) = -4 \frac{1}{(s-2)^2} \].

- Simplify the Equation

Combine and rearrange the terms to isolate $ Y(s)$. This results in: \[ (s^2 - 4s)Y(s) - 1 = -\frac{4}{(s-2)^2}. \]

- Solve for Y(s)

Isolate $ Y(s)$ by adding 1 to both sides and then dividing by $ s^2 - 4s $. We get: \[ Y(s) = -\frac{4}{(s-2)^2(s^2-4s)} + \frac{1}{s^2-4s}.\]

- Partial Fraction Decomposition

Next, perform partial fraction decomposition to simplify $ Y(s)$. For $ -\frac{4}{(s-2)^2(s^2-4s)}, $ rewrite it as: \[ -\frac{4}{(s-2)^2 s(s-4)} = -\frac{4}{s(s-2)^2(s-4)} = A/s + B/(s-2) + C/(s-2)^2 + D/(s-4). \].

- Find Constants in Partial Fraction decomposition

Equate and solve for constants $ A, B, C, D$ to solve for $ Y(s)$.

- Perform Inverse Laplace Transform

Find the inverse Laplace transform of $ Y(s) $ to get $ y(t) $. Use standard Laplace transform pairs and properties to revert to the time domain.

- Simplify the Solution

Simplify and combine all resulting terms from the inverse Laplace transform to write the final solution for $ y(t)$.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equations

A differential equation involves derivatives of a function. When we have a second-order differential equation like the one in the given problem, it includes the second derivative of the function. The equation given is: $ y'' - 4y' = -4te^{2t} $. Our goal is to solve for $ y(t) $, which means finding a function that satisfies this equation through all provided initial conditions.

Initial Conditions

Initial conditions are values that the solution of a differential equation must satisfy at a particular point, usually at $ t = 0 $. For instance, in our problem, we have these initial conditions:

$ y(0) = 0 $
$ y'(0) = 1 $

These conditions help us tailor our general solution specifically to the problem and are vital to simplify and solve the transformed equations correctly.

Inverse Laplace Transform

The inverse Laplace transform is the process of transforming a function from the Laplace (frequency) domain back to the time domain. In our problem, we have to invert the expression for $ Y(s) $ obtained after partial fraction decomposition. We use tables of standard Laplace transforms to assist us in identifying standard pairs. By applying these pairs, we achieve the original time-domain function $ y(t) $. Equations like: e.g., $ \frac{1}{s - a} \rightarrow e^{at} $ , $ \frac{1}{(s-a)^2} \rightarrow t e^{at} $, are used to get back function in the time domain.

Partial Fraction Decomposition

Partial fraction decomposition is a method that simplifies complex fractions into simpler terms, making them easier to handle in inverse Laplace transforms. For instance, for a function like: $ -\frac{4}{(s-2)^2(s^2-4s)} $ we perform partial fraction decomposition to break it into a sum of simpler fractions: $ \frac{A}{s} + \frac{B}{s-2} + \frac{C}{(s-2)^2} + \frac{D}{s-4} $. Identifying the constants $ A, B, C, $ and $ D $ allow us to then find the inverse Laplace transform of each term separately. This method is crucial for working through more complicated Laplace transforms.