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.