Question

If \(y_{1}\) and \(y_{2}\) are linearly independent solutions of \(t^{2} y^{\prime \prime}-2 y^{\prime}+(3+t) y=0\) and if \(W\left(y_{1}, y_{2}\right)(2)=3,\) find the value of \(W\left(y_{1}, y_{2}\right)(4)\)

Short answer

Based on the given second-order linear differential equation and Abel's theorem, we find that the value of the Wronskian W(y_1, y_2)(4) is equal to 3/4.

Step-by-step solution

Define the given differential equation and Wronskian

We are given the second-order linear differential equation: $$t^{2} y^{\prime \prime}-2 t y^{\prime}+(3+t) y=0.$$ And we know that \(y_1(t)\) and \(y_2(t)\) are linearly independent solutions of this equation. The Wronskian \(W(y_1, y_2)\) is defined as: $$W(y_1, y_2)(t) = y_1(t)y_2'(t) - y_1'(t)y_2(t).$$

Applying Abel's theorem to the differential equation

Abel's theorem states that for a second-order linear homogeneous differential equation, $$p(t)y'' + q(t)y' + r(t)y = 0, $$ the Wronskian can be found by: $$W(y_1, y_2)(t) = W(y_1, y_2)(t_0) \cdot e^{-\int_{t_0}^{t} \frac{q(s)}{p(s)} ds}.$$ In our case, \(p(t) = t^2\), \(q(t) = -2t\), and \(r(t) = 3+t\). So, the expression in the integral becomes, \(\frac{-2t}{t^2} = -\frac{2}{t}\).

Calculate the integral in Abel's theorem

We need to calculate the integral $$-\int_{t_0}^{t} \frac{2}{s} ds$$. This evaluates to: $$-2(\ln(t) - \ln(t_0)) = -2(\ln(\frac{t}{t_0}))$$.

Apply Abel's theorem to find the Wronskian for any t

Now, applying Abel's theorem with the calculated integral, we have: $$W(y_1, y_2)(t) = W(y_1, y_2)(t_0) \cdot e^{-2(\ln(\frac{t}{t_0}))}$$. We are given that \(W(y_1, y_2)(2) = 3\). Let's substitute \(t_0=2\) and \(t=4\) to find \(W(y_1, y_2)(4)\).

Find the value of the Wronskian at t = 4

Using \(t_0=2\), \(t=4\), and \(W(y_1, y_2)(2) = 3\), we get: $$W(y_1, y_2)(4) = 3 \cdot e^{-2(\ln(\frac{4}{2}))} = 3 \cdot e^{-2(\ln(2))} = 3 \cdot \frac{1}{4}.$$ Finally, we have: $$W(y_1, y_2)(4) = \frac{3}{4}.$$