Show that $F(f)(t) = t^2 + \frac{t}{3}f(t) + \frac{1}{5}\int_0^t e^uf(u) du$ is a contraction on $(C[0, 1), d_u)$.
Deduce that the differential equation $(15 − 5t)\frac{df}{dt} = (5 + 3e^{t})f + 30t$ has a unique solution in $C[0, 1]$.
Answer I have completed the first part and shown that $F$ is a contraction on $C[0, 1]$. Then by the contraction mapping theorem there exists a fixed point $f \in C[0, 1]$ such that $F(f)(t) = f(t)$
So I have $f(t) = t^2 + \frac{t}{3}f(t) + \frac{1}{5}\int_0^t e^uf(u) du$
and if I rearrange and differentiate I get $(15 − 5t)\frac{df}{dt} = (5 + 3e^{t})f + 30t$
So I know $f$ is a solution to the differential equation. But I have to now show it is the only solution. Normally that is done by taking another function, $g$, from $C[0, 1]$ that satisfies the equation and showing that it must equal $f$. However I am unable to make that work here.
And in fact, I don't understand how there can be a unique solution anyway when we are given no initial condition such as $x(0) = 1$...and then if we had $f(t) = t^2 + \frac{t}{3}f(t) + \frac{1}{5}\int_0^t e^uf(u)du$ + K, where $K$ is any constant, this will also satisfy the equation. So is this a trick question and the differential equation doesn't ac