Find the value of $W(y_1,y_2)(5)$.

Question

I have to solve the following exercise:

If $y_1$ and $y_2$ are linearly independent solutions of $ty'' + 2y' + te^ty = 0$ and if $W(y_1,y_2)(1) = 2$, find the value of $W(y_1,y_2)(5)$. ($W(y_1,y_2)(t_0)$ denotes the value of the Wronskian at $t = t_0$.)

I know that if $y_1$ and $y_2$ are linearly independent solutions of $$y'' + p(t)y' + q(t)y = 0$$ then the Wronskian is given by $W(y_1,y_2)(t_0) = c\exp[-\int p(t)dt]$. So I think I need to rewrite the original differential equation such that $y''$ has a coefficient of $1$.

Since I can't think of a way to do that now, I looked at the regular expression of the Wronskian: $W(y_1,y_2)(1) = y_1(1)y_2'(1) - y_1'(1)y_2(1) =2$. I tried to find more about this expression by looking at the differential equation at $t = 1$; $y''+ 2y' + ey = 0$, but this doesn't really get me anywhere either.

So my question is: How do I find the value of $W(y_1,y_2)(5)$?

Answer 1

You are right in that you need to rewrite the equation so that y′′ has a coefficient of 1.

But I think it is important that you know why. The thing is: if you write the explicit expression for the Wronskian, and then derive it with respect to time, you get a multiple of the same expression. That is: the time derivative of the Wronskian depends only on the Wronskian, and not on the separate values of y1(t) and y2(t). Let's see:

$ W(y_1(t), y_2(t)) = y_1(t) y'_2(t) - y_2(t) y'_1(t) $

$ W'(y_1(t), y_2(t)) = y_1(t) y''_2(t) - y_2(t) y''_1(t) $

Then you can try to use the original equation: $ ty′′+2y′+te^ty=0 $. For that, just multiply both sides of the Wronskian derivative by t:

$ tW'(y_1(t), y_2(t)) = y_1(t)(t y''_2(t)) - y_2(t) (t y''_1(t)) $

$ tW'(y_1(t), y_2(t)) = y_1(t)(-2y′_2-te^ty_2) - y_2(t) (-2y′_1-te^ty_1) $

$ tW'(y_1(t), y_2(t)) = -2 y_1(t) y′_2 + 2 y_2(t) y′_1 $

$ tW'(y_1(t), y_2(t)) = -2 W(y_1(t), y_2(t)) $

Now you have a differential equation for W only, which determines the value of W as a function W(t) of time. That is: the time derivative of W depends only on the value of W at each time step, not on it's expression in terms of y1 and y2. Then, you can solve this equation for W(t):

$ tW'(t) = -2 W(t) $

$ W'(t) = -(\frac{2}{t}) W(t) $

$ \frac{d}{dt} log W(t) = -2/t $

$ log W(t) = -2 log t + C $

$ W(t) = Ae^{-2 log t} $

$ W(t) = \frac{A}{t²} $

Then, if you know that W(1) = 2, you have A = 2, and

$ W(t) = \frac{2}{t^2} $

And you have the value of W for any time.

Of course, if you divided the whole equation by t and used $ W(t) = cexp[−∫p(t)dt] $, you would have the same result. I just made explicit the steps you need to get to that expression, but in a way you can directly apply the reasoning without worrying about the original equation's form.