How do I evaluate the Wronskian for this equation

Question

Martin Braun - Differential equations and their applications Chapter 2.1 p.137

Let $y_1,y_2$ be solutions of Bessel's equation $$t^2y'' + ty' + (t^2-n^2)y=0$$ on the interval $(0,\infty)$ with $y_1(1)=1,y_1'(1)=0,y_2(1)=0,y_2'(1)=1$. Compute the Wronskian $W[y_1,y_2](t)$.

The only thing I know is Wronskian is either zero of never a zero. But the Wronskian at $1$ in the exercise is $1\neq0$. How do I evaluate this?

Answer 1

The Wronskian $W$ of an equation of the form $$ y'' + py'+qy=0 $$ satisfies $$ W'= -pW. $$ In your case, you have $$ p(t)=\frac{1}{t}, $$ so you have $$ W(t) = W(1) \exp{\left( -\int_1^t ds/s \right)} = W(1) e^{-\log{t}} = \frac{W(1)}{t}, $$ and $W(1)=1$, as you note, so $W(t)=1/t$.

Answer 2

Hint: Start with the equation of evolution of the Wronskian of an ode written in the form $y^{\prime\prime}(t)=p(t)y'(t)+q(t)y(t)$

$$W'(t)=p(t)W(t)=-\frac{1}{t}W(t)$$

That can be obtained by just deriving

$$W(t)=\begin{vmatrix} y_1(t) & y_2(t)\\ y'_1(t) & y'_2(t) \end{vmatrix}$$

and keep in mind $W(1)=1$