Question

Show that if \(p\) is differentiable and \(p(t)>0,\) then the Wronskian \(W(t)\) of two solutions of \(\left[p(t) y^{\prime}\right]^{\prime}+q(t) y=0\) is \(W(t)=c / p(t),\) where \(c\) is a constant.

Short answer

Question: Show that for a second-order linear differential equation of the form \(\left[p(t) y^{\prime}\right]^{\prime}+q(t) y=0\), where \(p(t)>0\) and \(p(t)\) is differentiable, the Wronskian of two solutions is equal to a constant \(c\) divided by the function \(p(t)\). Answer: The Wronskian of two solutions to the given differential equation is \(W(t) = \frac{c}{p(t)}\), where \(c\) is a constant.

Step-by-step solution

Write down the Wronskian formula

Recall the Wronskian for two functions \(u(t)\) and \(v(t)\) is defined as \(W(t) = u(t)v'(t) - v(t)u'(t)\).

Find the expressions for \(u'(t)\) and \(v'(t)\) using the given differential equation

We can rewrite the given differential equation \(\left[p(t) y^{\prime}\right]^{\prime}+q(t) y=0\) as \(p(t) y^{\prime\prime} + p'(t) y^{\prime} + q(t) y = 0\). We have two solutions to this equation, \(u(t)\) and \(v(t)\). Plug these solutions into the rewritten equation: For \(u(t)\): \(p(t)u''(t) + p'(t)u'(t) + q(t)u(t) = 0\). For \(v(t)\): \(p(t)v''(t) + p'(t)v'(t) + q(t)v(t) = 0\).

Multiply the equations in Step 2 by \(v(t)\) and \(u(t)\), respectively, and subtract the two resulting equations

Multiply the first equation in Step 2 by \(v(t)\) and the second equation by \(u(t)\): \(v(t)p(t)u''(t) + v(t)p'(t)u'(t) + v(t)q(t)u(t) = 0\). \(u(t)p(t)v''(t) + u(t)p'(t)v'(t) + u(t)q(t)v(t) = 0\). Subtract the second equation from the first equation: \(v(t)p(t)u''(t) - u(t)p(t)v''(t) = u(t)p'(t)v'(t) - v(t)p'(t)u'(t)\).

Express the Wronskian using the result from Step 3

Note that the left-hand side of the equation in Step 3 is equal to \([v(t)u''(t) - u(t)v''(t)]p(t)\). The right-hand side is equal to \([u(t)v'(t) - v(t)u'(t)]p'(t)\). So we have: \(W'(t)p(t) = W(t)p'(t)\).

Integrate both sides of the equation with respect to \(t\)

To find the general solution of \(W'(t)p(t) = W(t)p'(t)\), we will integrate both sides with respect to \(t\). Recall that \(\frac{d}{dt} \left( \frac{W(t)}{p(t)} \right) = \frac{W'(t)p(t) - W(t)p'(t)}{p(t)^2}\). Thus, we can rewrite the equation as: \(\frac{d}{dt} \left( \frac{W(t)}{p(t)} \right) = 0\). Integrate both sides with respect to \(t\): \(\int \frac{d}{dt} \left( \frac{W(t)}{p(t)} \right) dt = \int 0 dt\).

Solve the integration and find the general solution for \(W(t)\)

Apply the fundamental theorem of calculus to the left-hand side and integrate the right-hand side directly: \(\frac{W(t)}{p(t)} = c\). Multiply both sides by \(p(t)\) to solve for the Wronskian: \(W(t) = c / p(t)\). This is the desired result. We have shown that the Wronskian of two solutions to the given differential equation is \(W(t) = \frac{c}{p(t)}\), where \(c\) is a constant.

Key concepts

Wronskian

The Wronskian is a determinant used in solving systems of linear differential equations to determine whether the solutions are linearly independent. For two functions, \( u(t) \) and \( v(t) \), the Wronskian \( W(t) \) is expressed as \( W(t) = u(t)v'(t) - v(t)u'(t) \). It's a measure of how their derivatives vary with respect to each other.

In the context of the problem, analyzing the Wronskian helps us understand how two particular solutions of the given differential equation behave. A nonzero Wronskian indicates that the solutions are linearly independent, which is a critical fact when proving the uniqueness and existence of solutions for differential equations.

Differentiability

Differentiability refers to the property of a function that allows it to be differentiated, which means a function \( f(t) \) has a derivative at every point in its domain. A differentiable function is necessarily continuous, but the converse is not true

In differential equations such as in this exercise, the concept of differentiability ensures that the functions we work with can be meaningfully manipulated. The differentiability of \( p(t) \) is essential in finding the derivatives needed to express the Wronskian and solve for \( W(t) = c / p(t) \). This guarantees that during calculations, operations like derivatives and integrations are performed without inconsistencies.

Integration

Integration is a fundamental concept in calculus that is essentially the inverse process of differentiation. It calculates the area under a curve or the accumulation of quantities, and is expressed in terms of integrals. In solving differential equations, integration can be used to find general solutions from differential equations that express relationships between functions and their derivatives.

In the step-by-step solution, integrating \( \frac{d}{dt} \left( \frac{W(t)}{p(t)} \right) = 0 \) involves finding a constant of integration, which leads to the expression \( \frac{W(t)}{p(t)} = c \). This equation represents a constant relationship that defines the Wronskian of the two solutions. The process of integration here is crucial to conclude that \( W(t) = c / p(t) \).

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the concepts of differentiation and integration, stating that differentiation and integration are inverse processes. It consists of two main parts: one relating antiderivatives to definite integrals, and the other asserting that the integral of a derivative results in the original function up to a constant.

In this problem, when integrating \( \frac{d}{dt} \left( \frac{W(t)}{p(t)} \right) = 0 \), the theorem allows us to assert that the integral of the derivative yields \( \frac{W(t)}{p(t)} \) as a constant \( c \). By restructuring the problem with this principle, the solution elegantly shows that the Wronskian \( W(t) \) is invariantly proportional to \( 1/p(t) \), demonstrating the power of the Fundamental Theorem of Calculus in providing solutions to complex differential equations.