Given $W[f,g](t) = t\sin(t)$ find $W[f+g,f-g]$

Question

If the $W[f,g](t) = t\sin(t)$, find $W[f+g,f-g]$. $f$ and $g$ are not given, I wrote down the original equation $fg'-gf'$ to compare to $f+g(f-g)-f-g(f+g)'= t\sin(t)$. I am not sure how to approach this problem.

Answer 1

Start with $W[f+g,f-g]$ and expand it to get

$$\begin{align} W[f+g,f-g]&=(f+g)(f-g)'-(f-g)(f+g)'\\ &=ff'-(fg'-gf')-gg'-(ff'+(fg'-gf')-gg')\\ &=-2(fg'-gf') \end{align}$$

and hence $fg'-gf'$ equals $W[f,g]$, as you already mentioned, we get

$$W[f+g,f-g]=-2W[f,g]=-2t\sin t$$

Answer 2

The Wronskian $$ W(f, g) = \begin{vmatrix} f & g \\ f' & g'\end{vmatrix} $$ “inherits” some properties of the determinant:

• It is linear in each argument.
• It is anti-symmetric, i.e. it changes the sign if two arguments are swapped.

(Those properties hold for Wronskians of arbitrary many functions, but we need it only for two functions here.)

Therefore $$ W(f+g, f-g) = W(f, f) + W(f, -g) + W(g, f) + W(g, -g) \\ = 0 - W(f, g) - W(f, g) + 0 = -2 W(f, g) \, . $$