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). $$ y^{\prime \prime}+(\sin t) y^{\prime}+t^{2} y=0 $$

Short answer

Answer: Yes, the given linear differential equation has a general solution of this form since it satisfies the conditions of Theorem 8.2(b).

Step-by-step solution

Identify p(t) and q(t) from the given equation

The given linear differential equation is: \(y''+(\sin t)y'+t^2y=0\). From this, we identify the given \(p(t) = \sin t\) and \(q(t) = t^2\).

Verify if the given p(t) and q(t) satisfy the conditions in Theorem 8.2(b)

According to the theorem, we need to check if \(p(-t) = -p(t)\) and \(q(-t) = q(t)\). Let's plug in the given p(t) and q(t) into these conditions: \(p(-t) = \sin(-t)\) and \(-p(t) = -\sin(t)\). Since \(\sin(-t) = -\sin(t)\), the first condition is satisfied. Now, let's check for the second condition: \(q(-t) = (-t)^2\) and \(q(t) = t^2\). Since \((-t)^2 = t^2\), the second condition is also satisfied.

Conclusion

Since both conditions in Theorem 8.2(b) are satisfied by the given p(t) and q(t), we can conclude that the given linear differential equation \(y''+(\sin t)y'+t^2y=0\) has a general solution of the form $$y(t)=c_1y_{\text{e}}(t)+c_2y_{\text{o}}(t),$$ where \(y_{\text{e}}(t)\) is an even solution and \(y_{\text{o}}(t)\) is an odd solution.

Key concepts

Linear Differential Equations

Linear differential equations are equations that involve the unknown function and its derivatives in a linear form. This means that the function and its derivatives are only multiplied by coefficients that can depend on the independent variable, but not the unknown function itself.
In a typical linear differential equation, like the one in our exercise, all terms involving the function and its derivatives are added together to sum up to zero or another function. For example, the equation \(y'' + (\sin t)y' + t^2 y = 0\) is a second-order linear differential equation because it involves the second derivative of \(y\), the first derivative \(y'\), and the function \(y\) itself, with coefficients that depend on \(t\).
These equations are crucial in modeling many real-world phenomena such as population growth, circuit designs, and mechanical vibrations. Understanding the structure and components of these equations helps in predicting the system’s behavior under various conditions.

General Solution

The general solution of a differential equation represents all possible solutions that satisfy the equation. It typically contains arbitrary constants, which accommodate the initial or boundary conditions of the problem.
In the context of our exercise, the solution to the differential equation \(y'' + p(t)y' + q(t)y = 0\) is expressed as \(y(t) = c_1y_e(t) + c_2y_o(t)\). Here, \(c_1\) and \(c_2\) are arbitrary constants. They allow for the flexibility needed to match initial conditions or any specific setup.
Even solutions, \(y_e(t)\), and odd solutions, \(y_o(t)\), embody the symmetry of the problem. Solutions are combined linearly to form the general solution, indicating that any combination of \(y_e\) and \(y_o\) is also a solution. This property is a cornerstone of linear systems.

Even and Odd Functions

In mathematics, even and odd functions are concepts related to symmetry. An even function satisfies the condition \(f(-t) = f(t)\). Graphically, even functions are symmetric with respect to the y-axis.
An odd function, on the other hand, satisfies the condition \(f(-t) = -f(t)\). These functions exhibit rotational symmetry around the origin.
In our exercise, the differential equation's solutions can have even and odd properties. This means the equation allows for solutions that mirror these symmetry characteristics. The use of even and odd functions in programs like solving differential equations can simplify the resolution process. These properties can help identify and verify the correct forms of solutions during calculations.