Question

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

Short answer

The solution is \( y(t) = 4t^2 e^{2t} + e^{2t} \).

Step-by-step solution

Take the Laplace transform of both sides

Apply the Laplace transform to both sides of the differential equation \[ \mathcal{L}\{y''(t) - 4y'(t)\} = \mathcal{L}\{-4t e^{2t}\} \]

Use Laplace transform properties on the left side

Use the properties of the Laplace transform: \[ \mathcal{L}\{y''(t)\} = s^2Y(s) - sy(0) - y'(0) \] \[ \mathcal{L}\{y'(t)\} = sY(s) - y(0) \] Given \( y(0) = 0 \) and \( y'(0)=1 \), substitute these into the transforms: \[ s^2Y(s) - 1 - 4(sY(s)) = \mathcal{L}\{-4t e^{2t}\} \]

Simplify the equation

Simplify the left-hand side of the equation: \[ s^2Y(s) - 4sY(s) - 1 = \mathcal{L}\{-4t e^{2t}\} \] Combine like terms: \[ (s^2 - 4s)Y(s) - 1 = \mathcal{L}\{-4t e^{2t}\} \]

Take the Laplace transform on the right side

Use the Laplace transform property for \(-4t e^{2t}\): \[ \mathcal{L}\{t e^{at}\} = \frac{1}{(s-a)^2} \] Therefore, \[ \mathcal{L}\{-4t e^{2t}\} = -4 \cdot \frac{1}{(s-2)^2} = -\frac{4}{(s-2)^2} \]

Combine both sides and solve for Y(s)

Plug the right-hand side result into the equation: \[ (s^2 - 4s)Y(s) - 1 = -\frac{4}{(s-2)^2} \] Move \(-1\) to the right side: \[ (s^2 - 4s)Y(s) = -\frac{4}{(s-2)^2} + 1 \]

Solve for Y(s)

Isolate \(Y(s)\): \[ Y(s) = \frac{-\frac{4}{(s-2)^2} + 1}{s^2 - 4s} \] Simplify the numerator: \[ Y(s) = \frac{-4 + (s-2)^2}{(s-2)^2 (s(s-4))} \]

Partial fraction decomposition

Break down \(Y(s)\) into partial fractions: \[ Y(s) = \frac{4}{(s-2)^3} + \frac{1}{(s-2)} \]

Inverse Laplace Transform

Use the inverse Laplace transform to return to the time domain: \[ y(t) = \mathcal{L}^{-1}\{Y(s)\} = \mathcal{L}^{-1}\{\frac{4}{(s-2)^3}\} + \mathcal{L}^{-1}\{\frac{1}{s-2}\} \] Using known transforms: \[ \frac{4}{(s-2)^3} \rightarrow 4t^2 e^{2t} \] and \( \frac{1}{s-2} \rightarrow e^{2t} \)

Write the final solution

Combine all the terms to find the final solution: \[ y(t) = 4t^2 e^{2t} + e^{2t} \] Simplify if possible.

Key concepts

Differential Equations

A differential equation is a mathematical equation that relates a function with its derivatives. In simpler terms, it defines how the function changes along with its rate of change. In the given exercise, we are tasked with solving the differential equation: \[ y^{\top} - 4 y^{'} = -4 t e^{2 t} \] This is a second-order differential equation because it involves the second derivative of the function y(t). Solving such equations often requires transforming them, simplifying them, and reverting back to the original variable.

Initial Conditions

Initial conditions are essential for uniquely determining the solution to a differential equation. They specify the values of the function and its derivatives at a particular point, usually t = 0. In our problem, the initial conditions are given as:

• \( y(0) = 0 \)
• \( y'(0) = 1 \)

These conditions allow us to plug specific values into our Laplace transforms, enabling us to simplify the equation and solve for Y(s) effectively.

Partial Fraction Decomposition

Partial fraction decomposition is a method used to break down complex rational expressions into simpler fractions. This makes taking the inverse Laplace transform more straightforward. In the exercise, we encounter an expression for \( Y(s) \) that needs to be decomposed for easier handling: \[ Y(s) = \frac{-\frac{4}{(s-2)^2} + 1}{s^2 - 4s} \] We decompose it into partial fractions to find: \[ Y(s) = \frac{4}{(s-2)^3} + \frac{1}{(s-2)} \] This transformation simplifies the process of finding the inverse Laplace transform.

Inverse Laplace Transform

The inverse Laplace transform is used to convert a function in the Laplace domain back to the time domain. After partial fraction decomposition, we apply the inverse Laplace transform to each term:

• \( \frac{4}{(s-2)^3} \rightarrow 4t^2 e^{2t} \)
• \( \frac{1}{s-2} \rightarrow e^{2t} \)

Combining these, we get the final solution to the differential equation in time domain: \[ y(t) = 4t^2 e^{2t} + e^{2t} \] This process effectively resolves the initial problem by reverting the transformed function to its original form.