Question

For the HSOLDE \(y^{\prime \prime}+p y^{\prime}+q y=0\), show that $$ p=-\frac{f_{1} f_{2}^{\prime \prime}-f_{2} f_{1}^{\prime \prime}}{W\left(f_{1}, f_{2}\right)} \text { and } q=\frac{f_{1}^{\prime} f_{2}^{\prime \prime}-f_{2}^{\prime} f_{1}^{\prime \prime}}{W\left(f_{1}, f_{2}\right)} . $$ Thus, knowing two solutions of an HSOLDE allows us to reconstruct the DE.

Short answer

The functions \(p\) and \(q\) are expressed by \(p=-\frac{f_{1} f_{2}^{\prime \prime}-f_{2} f_{1}^{\prime \prime}}{W\left(f_{1}, f_{2}\right)}\) and \(q=\frac{f_{1}^{\prime} f_{2}^{\prime \prime}-f_{2}^{\prime} f_{1}^{\prime \prime}}{W\left(f_{1}, f_{2}\right)}\), respectively, where \(W(f_1, f_2)\) is the Wronskian of the two solutions \(f_{1}\) and \(f_{2}\) to the homogeneous second-order linear differential equation (HSOLDE).

Step-by-step solution

Compute the Derivatives and the Wronskian

It's necessary to compute the first, second and mixed second derivatives of the functions \(f_{1}\) and \(f_{2}\). Verify that the Wronskian \(W(f_{1}, f_{2}) = f_{1} f_{2}^{\prime} - f_{2}f_{1}^{\prime} \neq 0\). This condition is a requirement for the given solutions.

Compute the Formulas for p and q

For \(p\), multiply the HSOLDE by \(f_{2}\) and subtract this from the product of the function \(f_{1}\) and the derivative of \(f_{2}\). The result should be equal to \(p\) times the Wronskian. Solving for \(p\), it can be obtained that \(p=-\frac{f_{1} f_{2}^{\prime \prime}-f_{2} f_{1}^{\prime \prime}}{W\left(f_{1}, f_{2}\right)}\). Similarly, for \(q\), multiply the HSOLDE by \(f_{2}\) and add it to the product of \(f_{1}\) and the second derivative of \(f_{2}\). The result should be equal to \(q\) times the Wronskian. Solving for \(q\), it can be obtained that \(q=\frac{f_{1}^{\prime} f_{2}^{\prime \prime}-f_{2}^{\prime} f_{1}^{\prime \prime}}{W\left(f_{1}, f_{2}\right)}\).

Validate the Solutions

Plugging these relations for \(p\) and \(q\) into the given HSOLDE will verify the correctness of the found expressions. It should obtain the original formula. If it checks out, then you have shown the requested properties.

Key concepts

Wronskian determinant

In the study of higher-order linear differential equations, the Wronskian determinant plays a vital role. It is used to determine whether a set of solutions is linearly independent. When dealing with a system of differential equations, knowing if your solutions are independent ensures you can construct a general solution for the problem.
To compute the Wronskian for two functions, say, \(f_1\) and \(f_2\), you set it up as \(W(f_1, f_2) = f_1 f_2^{\prime} - f_2 f_1^{\prime}\). It is vital that this Wronskian is non-zero.

• If \(W(f_1, f_2) eq 0\), the solutions are independent, indicating numerous general solutions for the differential equation.
• If \(W(f_1, f_2) = 0\), the solutions might be dependent, and further investigation or a different approach might be needed.

Checking the Wronskian condition ensures you have a solid foundation to work with the differential equation you're trying to solve.

Differential equation solutions

Differential equation solutions are the specific functions that satisfy a given differential equation, like the higher-order linear differential equation mentioned in the step-by-step solution. These solutions can often be represented as function combinations that include exponential, sine, and cosine terms.
For the task at hand, knowing two solutions \(f_1\) and \(f_2\) allows one to delve deeper. The relationships derived for \(p\) and \(q\) seem complex, yet utilizing the Wronskian simplifies identifying them. By multiplying parts of the equation by given functions and performing arithmetic, \(p\) and \(q\) were isolated:

• \(p=-\frac{f_{1} f_{2}^{\prime \prime}-f_{2} f_{1}^{\prime \prime}}{W(f_1, f_2)}\).
• \(q=\frac{f_{1}^{\prime} f_{2}^{\prime \prime}-f_{2}^{\prime} f_{1}^{\prime \prime}}{W(f_1, f_2)}\).

By following these steps, students can appreciate solver strategies and reconstruct the differential equation contextually from its solutions.

Regular differential equations

In the context of higher-order linear differential equations, the term "regular" means that the equation behaves well under certain conditions and can be solved using established methods. In these computations, regularity implies the absence of singularities, where coefficients could potentially become undefined or infinite.
When an equation is regular, it implies all functions and parameters involve a smooth behavior that guarantees a unique, well-behaved set of solutions:

• Regular equations usually ensure better conditions for initial value or boundary value problems, allowing traditional techniques to be used with greater confidence.
• These equations allow useful properties like Wronskians and solutions consistency to be applied directly, without worrying about special cases or exceptions.

Understanding the concept of regularity equips students with better tools to tackle complex differential equations, acknowledging predictable behavior and consistency.