Question

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

Short answer

y(t) = 2 - te^{2t}

Step-by-step solution

Take the Laplace Transform of both sides

First, apply the Laplace transform to both sides of the differential equation \[y'' - 4y' + 4y = 4\] Using the properties of Laplace transforms, we get: \[ L\big\{y''\big\} - 4L\big\{y'\big\} + 4L\big\{y\big\} = L\big\{4\big\} \]\[ s^2Y(s) - sy(0) - y'(0) - 4(sY(s) - y(0)) + 4Y(s) = \frac{4}{s}\]Substitute the initial conditions \(y(0) = 0\) and \(y'(0) = -2\): \[ s^2Y(s) - 0 + 2 - 4(sY(s) - 0) + 4Y(s) = \frac{4}{s}\]

Simplify and solve for \(Y(s)\)

Combine terms to isolate \(Y(s)\): \[ s^2Y(s) + 2 - 4sY(s) + 4Y(s) = \frac{4}{s}\] Factor out \(Y(s)\): \[ (s^2 - 4s + 4)Y(s) + 2 = \frac{4}{s}\] Subtract 2 from both sides: \[ (s^2 - 4s + 4)Y(s) = \frac{4}{s} - 2 \] Note that \(s^2 - 4s + 4 = (s-2)^2\), so: \[ Y(s)(s-2)^2 = \frac{4}{s} - 2 \]

Further simplify to solve for \(Y(s)\)

Multiply the right side to combine fractions: \[ (s-2)^2 Y(s) = \frac{4 - 2s}{s} \] Divide both sides by \((s-2)^2\): \[ Y(s) = \frac{4 - 2s}{s(s-2)^2} \]

Perform partial fraction decomposition

Decompose \( Y(s) = \frac{4 - 2s}{s(s-2)^2}\) into partial fractions: \[ \frac{4 - 2s}{s(s-2)^2} = \frac{A}{s} + \frac{B}{s-2} + \frac{C}{(s-2)^2} \] Using an algebraic method to find constants \(A\), \(B\), and \(C\), we determine: \[ Y(s) = \frac{2}{s} + \frac{-1}{(s-2)^2}\]

Apply the inverse Laplace transform

Now, take the inverse Laplace transform to find \(y(t)\). The transform pairs are: \[L^{-1}\bigg\{\frac{2}{s}\bigg\} = 2\] \[L^{-1}\bigg\{\frac{-1}{(s-2)^2}\bigg\} = -te^{2t}\] Therefore, \[ y(t) = 2 - te^{2t} \]

Key concepts

differential equations

Differential equations are mathematical equations that involve derivatives of a function. They describe how a particular quantity changes over time. In this exercise, we deal with a second-order linear differential equation: \[y'' - 4y' + 4y = 4\] where \(y''\) is the second derivative of \(y\) with respect to time, and \(y'\) is the first derivative.
These equations are essential for modeling various real-world problems including motion, electricity, and fluid dynamics.
To solve this differential equation, we used the Laplace Transform, a powerful technique often used to convert differential equations into algebraic equations which are simpler to handle.

initial conditions

Initial conditions are specific values given at the start of the problem, which help in finding a unique solution to the differential equation. For this problem, the initial conditions are given as:

• \(y(0) = 0\)
• \(y'(0) = -2\)

These initial conditions are plugged into the solution after applying the Laplace Transform to the differential equation. They help determine the constants in the resulting algebraic equation. Without these initial conditions, we would have an infinite number of solutions.

inverse Laplace transform

The inverse Laplace Transform is used to convert back from the transformed (algebraic) domain to the time domain. After solving the algebraic equation and expressing \(Y(s)\) as a function of \(s\), we use the inverse Laplace transform to find \(y(t)\).
In this exercise, we found that: \[Y(s) = \frac{2}{s} - \frac{1}{(s-2)^2}\] Using known inverse Laplace transform pairs, we have:

• \(L^{-1}\left\{\frac{2}{s}\right\} = 2\)
• \(L^{-1}\left\{\frac{-1}{(s-2)^2}\right\} = -te^{2t}\)

Adding these results gives the final solution: \[y(t) = 2 - te^{2t}\]

partial fraction decomposition

Partial fraction decomposition is a method used to break down complex rational functions into simpler fractions that are easier to invert using Laplace Transforms. In this exercise, we decomposed: \[\frac{4 - 2s}{s(s-2)^2}\] into simpler fractions. The goal is to express this as: \[\frac{A}{s} + \frac{B}{s-2} + \frac{C}{(s-2)^2}\] By finding constants \(A, B,\) and \(C\), we determined that: \[Y(s) = \frac{2}{s} + \frac{-1}{(s-2)^2}\] This decomposition allowed us to easily find the inverse Laplace transform of each term, leading to the solution \(y(t)\).