Wronskian of second order differential equation $ty'' − (t + 1)y' − y = 0$.

Question

Find a Wronskian of two solutions of $$ty'' − (t + 1)y' − y = 0, ~~~t > 0$$ provided $W[y_1, y_2](1) = 1$.

Answer: $W [y_1, y_2] (t) = te^{t−1}$

I am unsure of how they got the answer, am I supposed to use Abel's theorem?

When I do use Abel's theorem that is not the answer I get.

Can someone please explain this step by step?

Answer 1

$$ty′′ − (t + 1)y' − y = 0, t > 0$$

$$y′′= \frac {(t + 1)} t y' + \frac 1 t y $$

$$y′′= a(t)y' + b(t) y $$

Now remember that $$W(t)=e^{\int a(t)dt}$$ $$W(t)=e^{\int \frac {t+1}{t}}$$ $$W(t)=Ke^tt$$

Evaluate the integrale then find the value of the constant K since $W(1)=1$

Answer 2

Use Abel's theorem then for $p(t)=\dfrac{t+1}{t}$ $$W(t)=Ce^{-\int pdt}=Ce^{\int\frac{t+1}{t}dt}=Cte^t$$ with $W(1)=1$ you find $C=\dfrac1e$ so $W(t)=\dfrac1ete^t=te^{t-1}$.

Answer 3

For a good reference refer notes

Yes, we should be using Abel's theorem to calculate the Wronskian of the ODE.

It is useful since without knowing the solutions we are able to calculate the Wronskian of the ODE.

So,

$W(y_{1},y_{2})(t) = c e^{\int{\frac{t+1}{t}} dt} = cte^{t}$

and since $W(y_{1},y_{2})(1) = 1$ so $c = \frac{1}{e}$

thus the Wronskian of the above ODE is $W(y_{1},y_{2})(t) = \frac{1}{e} t e^{t}$.