Problem 18 - Calculus: Early Transcendentals

Question

Solve the initial value problem. $$y^{\prime \prime}+9 y=4 t, y(0)=0, y^{\prime}(0)=0$$

Step-by-Step Solution

Answer

The solution is \(y(t) = -\frac{4}{27} \sin(3t) + \frac{4}{9}t\).

Step 1: Solve the Homogeneous Equation

The homogeneous part is given by the equation \(y'' + 9y = 0\). This is a second-order linear homogeneous differential equation. The characteristic equation is \(r^2 + 9 = 0\). Solving for \(r\), we get \(r^2 = -9\), so \(r = \pm 3i\). The general solution for the homogeneous equation is \(y_h(t) = C_1 \cos(3t) + C_2 \sin(3t)\).

Step 2: Solve the Particular Solution

To solve the particular solution, let's assume a particular solution of the form \(y_p(t) = At + B\). Substitute \(y_p\) into the original equation \(y'' + 9y = 4t\). Calculating derivatives, \(y_p' = A\) and \(y_p'' = 0\). Substitute these into the equation: \(0 + 9(At + B) = 4t\). Simplify to get \(9At + 9B = 4t\). Equating coefficients for terms of \(t\), we have \(9A = 4\) (so \(A = \frac{4}{9}\)) and \(9B = 0\) (so \(B = 0\)). Therefore, the particular solution is \(y_p(t) = \frac{4}{9}t\).

Step 3: Combine Solutions

The general solution to the differential equation is the sum of the homogeneous solution and the particular solution: \(y(t) = y_h(t) + y_p(t) = C_1 \cos(3t) + C_2 \sin(3t) + \frac{4}{9}t\).

Step 4: Apply Initial Conditions

Given the initial conditions \(y(0) = 0\) and \(y'(0) = 0\), we can substitute \(t=0\) into the general solution: \(y(0) = C_1 \cos(0) + C_2 \sin(0) + \frac{4}{9}(0) = C_1 = 0\). For \(y'(0)\), first compute the derivative of the general solution: \(y'(t) = -3C_1 \sin(3t) + 3C_2 \cos(3t) + \frac{4}{9}\). Substitute \(t=0\): \(y'(0) = 3C_2 + \frac{4}{9} = 0\). Solving for \(C_2\), we find \(3C_2 = -\frac{4}{9}\) and thus \(C_2 = -\frac{4}{27}\).

Step 5: Write the Final Solution

Substitute \(C_1 = 0\) and \(C_2 = -\frac{4}{27}\) back into the general solution: \(y(t) = 0 \cdot \cos(3t) - \frac{4}{27} \sin(3t) + \frac{4}{9}t\). Thus, the solution to the initial value problem is \(y(t) = -\frac{4}{27} \sin(3t) + \frac{4}{9}t\).

Key Concepts

Initial Value Problem

An Initial Value Problem (IVP) in the context of differential equations involves finding a function that satisfies a differential equation and also meets specific initial conditions.
In the provided exercise, we start with the differential equation:\[ y'' + 9y = 4t, \]accompanied by the initial conditions:
\( y(0) = 0 \) and \( y'(0) = 0 \).
These initial conditions specify the value of the function and its derivative at \( t=0 \).
Solving an IVP means finding a particular solution that not only satisfies the differential equation but also the initial conditions.
This gives a unique solution tailored to the specific problem, unlike general solutions which have arbitrary constants.
IVPs are important in various real-world applications, including physics and engineering, as they provide precise solutions that model specific scenarios.

Homogeneous Equation

A homogeneous equation is characterized by the absence of a non-zero constant or forcing function on the right-hand side of the equation.
In our example, the homogeneous equation is:\[ y'' + 9y = 0. \]This type of equation is particularly useful because it provides the complimentary solution to the differential equation.
The structure of the general solution reflects the equation's roots, determined through its characteristic equation.
The solution to the homogeneous equation is critical, as it forms the basis for the complete solution to the differential equation.

• In our case, the solution is expressed using trigonometric functions: \( y_h(t) = C_1 \cos(3t) + C_2 \sin(3t) \).

These components align with the nature of the roots (imaginary in this instance), and are vital for building the complete solution to the differential equation.

Particular Solution

The particular solution of a differential equation accounts for any specific terms or sources present in the non-homogeneous part of the equation.
In our exercise, we aim to find a solution to:\[ y'' + 9y = 4t. \]The method of finding the particular solution typically involves assuming a form that mimics the non-homogeneous part, in this case, a linear polynomial due to the presence of the term \(4t\).

• An assumed form for \( y_p(t) \) is \( At + B \).
• Substituting, we set up equations to solve for \( A \) and \( B \) resulting in \( y_p(t) = \frac{4}{9}t \).

The particular solution complements the homogeneous solution, and together they form the general solution.

Characteristic Equation

The characteristic equation is a vital tool for solving linear differential equations, particularly homogeneous equations.
It is derived from the differential equation by substituting the trial solution involving exponential functions.
In our example, the characteristic equation arose from:\[ y'' + 9y = 0, \]which leads to:

• \( r^2 + 9 = 0 \).
• The roots \( r = \pm 3i \) reflect the nature of the solutions, indicating oscillatory behavior captured by sine and cosine functions.

This characteristic equation forms the foundation for finding the general solution to the homogeneous part, revealing critical insights about the nature of solutions.
Specifically, the imaginary roots indicate that the system's natural behavior is oscillatory, leading to solutions expressed in terms of trigonometric functions.