Question

Suppose a linear differential equation \(y^{\prime \prime}+p(t) y^{\prime}+q(t) y=0\) satisfies the hypotheses of Theorem \(8.2(\mathrm{~b})\), on the interval \(-\infty< t <\infty\). Then, by Exercise 32 , we can assume the general solution of \(y^{\prime \prime}+p(t) y^{\prime}+q(t) y=0\) has the form $$ y(t)=c_{1} y_{\mathrm{e}}(t)+c_{2} y_{\mathrm{o}}(t), $$ where \(y_{\mathrm{e}}(t)\) is an even solution of \(y^{\prime \prime}+p(t) y^{\prime}+q(t) y=0\) and \(y_{\mathrm{o}}(t)\) is an odd solution. In each of the following exercises, determine whether Theorem \(8.2(\mathrm{~b})\) can be used to guarantee that the given differential equation has a general solution of the form (9). If your answer is no, explain why the equation fails to satisfy the hypotheses of Theorem \(8.2\) (b). $$ y^{\prime \prime}+t^{2} y=0 $$

Short answer

Based on the analysis of the given differential equation \(y^{\prime \prime} + t^2y = 0\), we have determined that it does indeed satisfy the hypotheses of Theorem 8.2b. As a result, the theorem does guarantee that the differential equation has a general solution in the form: $$ y(t) = c_1 y_e(t) + c_2 y_o(t). $$ This is due to the coefficient functions, \(p(t)\) and \(q(t)\), being continuous and even, and the linear independence of the even and odd solutions, \(y_e(t)\) and \(y_o(t)\).

Step-by-step solution

Check if the coefficient functions are continuous and even

We have \(p(t)=0\) and \(q(t)=t^2\). Both functions are continuous for all \(t\) and both are even functions as \(p(-t)=0=p(t)\) and \(q(-t)=(-t)^2=t^2=q(t)\), confirming the first two hypotheses.

Inspect the given differential equation

The given differential equation is \(y^{\prime \prime} + t^2 y = 0\). We can rewrite it as: \(y^{\prime \prime} = -t^2 y\).

Find a pair of linearly independent solutions

Since \(p(t)\) and \(q(t)\) are both even functions, the given linear differential equation is symmetric. Hence, \(y_e(t)\) and \(y_o(t)\) should exist and form a linearly independent pair of solutions. Therefore, we can conclude that the given differential equation satisfies the hypotheses of Theorem 8.2b and has a general solution of the form: $$ y(t) = c_1 y_e(t) + c_2 y_o(t). $$

Key concepts

Even and Odd Functions

In mathematics, particularly in the context of functions, a function is said to be even if it satisfies the condition \( f(-x) = f(x) \) for all \( x \). Conversely, a function is odd if it meets the condition \( f(-x) = -f(x) \).
These properties are significant when dealing with solutions to linear differential equations. Why? Because the symmetry of even and odd functions can help simplify the process of finding solutions. For instance, in the problem \( y^{\prime \prime}+t^2 y=0 \), the functions \( p(t) = 0 \) and \( q(t) = t^2 \) are both even, meaning they remain unchanged when the sign of \( t \) is flipped.
This symmetry ensures that our solutions can be expressed as a combination of even and odd functions, offering a robust approach to figure out the general solution of the equation.

General Solution

The term "general solution" in the context of linear differential equations refers to a solution that encompasses all possible specific solutions of the differential equation.
In exercises like the one described, where the equation \( y^{\prime \prime}+p(t) y^{\prime}+q(t) y=0 \) is given, the general solution is expressed as a combination of two basic solutions. These solutions are often comprised of linear combinations of even and odd functions.
Specifically, the general solution takes the form:
\[ y(t) = c_1 y_e(t) + c_2 y_o(t) \]
Here, \( c_1 \) and \( c_2 \) are arbitrary constants, \( y_e(t) \) is an even function, and \( y_o(t) \) is an odd function. This structure allows for flexibility and a comprehensive coverage of all potential behaviors of the system dictated by the differential equation.

Continuity of Coefficient Functions

The continuity of the coefficient functions in a linear differential equation is crucial for ensuring the existence of its solutions over an interval.
For the equation \( y^{\prime \prime}+p(t) y^{\prime}+q(t) y=0 \), where \( p(t) \, \) and \( q(t) \, \) are coefficients, these functions should be continuous over the interval of consideration. The continuity condition guarantees that solutions behave nicely, without any jumps or discontinuities, which can complicate analysis and solution finding.
In the step-by-step solution, it was verified that both \( p(t)=0 \) and \( q(t)=t^2 \) are continuous across all real numbers. This is fundamental in applying Theorem 8.2b, ensuring that the symmetry properties of the differential equation can be utilized effectively to find a comprehensive general solution.

Linearly Independent Solutions

Linearly independent solutions are a cornerstone in solving homogeneous linear differential equations.
A set of functions is said to be linearly independent if no function in the set can be written as a linear combination of the others. In the context of the differential equation \( y^{\prime \prime}+t^2 y=0 \), finding two such solutions, \( y_e(t) \) and \( y_o(t) \), is essential.
Why? Because these solutions allow for the complete and robust formation of the general solution. If functions are linearly dependent, they essentially mirror each other in some respect, which means they fail to cover the entire solution space.
In simpler terms, having linearly independent solutions is like having DNA with a wide variety of traits — it ensures that every possible behavior dictated by the equation is accounted for, providing a vibrant and complete description of all potential solutions.