Question

Suppose \(p\) and \(q\) are continuous on \((a, b)\) and \(y_{1}\) and \(y_{2}\) are solutions of $$y^{\prime \prime}+p(x) y^{\prime}+q(x) y=0$$ on \((a, b) .\) Let $$z_{1}=\alpha y_{1}+\beta y_{2} \quad \text { and } \quad z_{2}=\gamma y_{1}+\delta y_{2}$$ where \(\alpha, \beta, \gamma,\) and \(\delta\) are constants. Show that if \(\left\\{z_{1}, z_{2}\right\\}\) is a fundamental set of solutions of \((\mathrm{A})\) on \((a, b)\) then so is \(\left\\{y_{1}, y_{2}\right\\}\).

Short answer

Since we have shown that \(W(y_1, y_2)\) is nonzero, it implies that \(y_1\) and \(y_2\) are linearly independent. Therefore, we can conclude that \(\{y_1, y_2\}\) forms a fundamental set of solutions for the given differential equation.

Step-by-step solution

Substituting \(z_1\) and \(z_2\) into the differential equation

First, let's substitute \(z_1\) and \(z_2\) into the given differential equation: $$y^{\prime\prime} + p(x)y^{\prime} + q(x)y = 0.$$ To find the first and second derivatives of \(z_1\) and \(z_2\), we will use the fact that \(y_1\) and \(y_2\) are solutions to the equation. This means their derivatives satisfy the equation as well. We have: $$z_1' = \alpha y_1' + \beta y_2',$$ $$z_1'' = \alpha y_1'' + \beta y_2'',$$ $$z_2' = \gamma y_1' + \delta y_2',$$ $$z_2'' = \gamma y_1'' + \delta y_2''.$$ -BEGIN

Verifying that \(z_1\) and \(z_2\) satisfy the differential equation

Now, we will substitute the derivatives of \(z_1\) and \(z_2\) in our original differential equation: For \(z_1:\) $$(\alpha y_1'' + \beta y_2'') + p(x)(\alpha y_1' + \beta y_2') + q(x)(\alpha y_1 + \beta y_2) = \alpha (y_1'' + p(x)y_1' + q(x)y_1) + \beta (y_2'' + p(x)y_2' + q(x)y_2) = 0.$$ For \(z_2:\) $$(\gamma y_1'' + \delta y_2'') + p(x)(\gamma y_1' + \delta y_2') + q(x)(\gamma y_1 + \delta y_2) = \gamma (y_1'' + p(x)y_1' + q(x)y_1) + \delta (y_2'' + p(x)y_2' + q(x)y_2) = 0.$$ Thus, both \(z_1\) and \(z_2\) are solutions of the given differential equation.

Calculate the Wronskian of \(y_1\) and \(y_2\)

To show that \(\{y_1, y_2\}\) is a fundamental set of solutions, we need to prove that they are linearly independent. One way to do this is by calculating their Wronskian \(W(y_1, y_2)\): $$W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} = y_1 y_2' - y_1' y_2.$$ -BEGIN

Calculate the Wronskian of \(z_1\) and \(z_2\)

Now, let's calculate the Wronskian of \(z_1\) and \(z_2\), which we know is nonzero since they form a fundamental set of solutions: $$W(z_1, z_2) = \begin{vmatrix} z_1 & z_2 \\ z_1' & z_2' \end{vmatrix} = z_1 z_2' - z_1' z_2 =( \alpha y_1 + \beta y_2)(\gamma y_1' + \delta y_2') - (\alpha y_1' + \beta y_2')(\gamma y_1 + \delta y_2).$$ -BEGIN

Relating the Wronskians of \(y_1\) and \(y_2\) to \(z_1\) and \(z_2\)

Let us simplify the above expression: $$W(z_1, z_2) = (\alpha\gamma(y_1 y_1') + \alpha\delta(y_1 y_2') + \beta\gamma(y_2 y_1') + \beta\delta(y_2 y_2')) - (\alpha\gamma(y_1 y_1') + \alpha\delta(y_1' y_2) + \beta\gamma(y_1' y_1) + \beta\delta(y_1' y_2')).$$ We can rewrite the Wronskian as: $$W(z_1, z_2) = \alpha\delta(y_1 y_2'-y_1'y_2) - \beta\gamma(y_1 y_2'-y_1'y_2) = (\alpha\delta - \beta\gamma)W(y_1,y_2).$$ Since \(W(z_1, z_2)\) is nonzero (because \(\{z_1, z_2\}\) form a fundamental set of solutions), it follows that \(W(y_1,y_2)\) is also nonzero. -BEGIN

Conclude that \(\{y_1, y_2\}\) forms a fundamental set of solutions

We have shown that \(W(y_1, y_2)\) is nonzero, which implies that \(y_1\) and \(y_2\) are linearly independent. Thus, we can conclude that \(\{y_1, y_2\}\) forms a fundamental set of solutions for the given differential equation.

Key concepts

Wronskian

Understanding the Wronskian is essential for students wrestling with differential equations. It's a practical tool used to determine whether a set of functions are linearly independent. If we have two functions, say, y1 and y2, their Wronskian is given by the determinant of a matrix with their values and derivatives:
\[W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \ y_1' & y_2' \end{vmatrix} = y_1 y_2' - y_1' y_2.\]
Imagine plotting y1 and y2 on a graph. When the area shaped by their curves and the x-axis doesn't collapse into a single line at any point in the interval, y1 and y2 are linearly independent. In simpler terms, their Wronskian is not zero, signifying that they do not mimic each other by a constant factor—this uniqueness is essential for constructing the general solution to a differential equation.

Linear Independence

Linear independence is the backbone of understanding solutions to differential equations. Similar to a basketball team where each player has a unique role, in a set of solutions, each function must contribute something new.
For functions y1 and y2, linear independence means we can't find constants 'c1' and 'c2' (not both zero) that would make c1y1 + c2y2 = 0 true for all values within an interval. When we calculate the Wronskian and find it's not zero, we've essentially confirmed that our functions y1 and y2 play different roles in the solution to our differential equation—that no function is redundant.

Continuous Coefficients

When we speak of continuous coefficients in the realm of differential equations, we're referring to functions that smoothly vary without any jumps or abrupt changes in an interval. In our context, p(x) and q(x) are these coefficients. Their continuous nature is vital as they represent the physical or mathematical conditions of the problem, ensuring that the solutions we derive from them (y1, y2) are applicable and behave nicely across the domain. This continuity ensures that our solutions make sense and can be analyzed using classical calculus techniques.

Boundary Value Problems

Lastly, boundary value problems (BVPs) are a type of differential equation where we know the solution at the boundaries of an interval. Imagine bending a pipe cleaner: knowing its shape at the ends affects its curvature everywhere else. Similarly, in BVPs, knowing the solution specifics at the start and end points (a and b) helps us figure out the potential solution path.
BVPs are pivotal in physics and engineering, where these endpoints often represent fixed conditions like temperature or pressure. Solving BVPs involves finding a function that satisfies the given differential equation and meets these boundary conditions—this can often be done using techniques that involve the Wronskian and principles of linear independence we've just discussed.