Impulse response of Linear time-varying system

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Suppose we are given the following system:

$ \begin{align*} \dot{x} &= cos(t)x + \frac{1}{t+1}u \\ y &= sin(t)x \end{align*} $

I'm trying to find the impulse response of this system g(t,$\tau$). My attempt: $\\$

First we need to find the state transition matrix which can ben found by seperation of variables in the first differential equation. This matrix turns out to be:

$\Phi(t,\tau)$ = $e^{sin(t-\tau)}$

\begin{align*} x(t) = e^{sin(t-t_0)}x(t_0) + \int_{t_0}^{t}e^{sin(t-\tau)}\frac{1}{\tau + 1} u(\tau) d\tau \end{align*}

when $u(\tau) = \delta(\tau)$ and initial conditions are zero then:

\begin{align*} g(t,\tau) = sin(t)e^{sin(t)} \end{align*}

Why this answer does not depend on $\tau$ which we normally expect. Do you spot any mistakes in this attempt?

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The integrating factor is just $e^{-\sin(t)}$. This gives the solution formula $$ x(t)=e^{\sin(t)-\sin(t_0)}x_0+\int_{t_0}^te^{\sin(t)-\sin(s)}\frac1{1+s}u(s)\,ds. $$ Now if $u(s)=\delta_\tau(s)=\delta_0(s-\tau)$, then $$ x(t)=e^{\sin(t)-\sin(t_0)}x_0+\theta(t-τ)e^{\sin(t)-\sin(\tau)}\frac1{1+\tau}, $$ with $θ$ for the Heaviside unit jump function.