Integration and derivative of composite function

Question

Consider a 1st order linear time-varying differential equation of the form: $$\frac{d y(t)}{dt} = \alpha(t) y(t)$$ where $\alpha(t)$ is a scalar function. The time-solution to the above equation is known to be: $$ y(t) = e^{\int_0^t \alpha(\tau)d\tau}y(0).$$ Now, consider the following scalar function $x(p(t),t)$, where $p(t)$ is a function of time. Assume that you are given that: $$\frac{d x(p(t),t)}{dt} = a(p(t)) x(p(t),t)~~~~~~~~~(*)$$ where $a(p(t))$ is a positive scalar function. I believe, the integration with respect to time is given by $$ x(p(t),t) = e^{\int_0^t a(p(\tau))d\tau} x(p(0),0) ~~~~~~~~(**)$$
However, if I want to take the total derivative we have: $$ \frac{d x(p(t),t)}{dt} = \frac{\partial x}{\partial p}\frac{dp}{dt}+\frac{\partial x}{\partial t}$$ $$ \frac{d x(p(t),t)}{dt} = x(p(t),t)\int_0^t \frac{\partial a(p(\tau))}{dp}d\tau \frac{dp(t)}{dt}+a(p(t))\cdot x(p(t),t)$$ But the above equation is obviously not equal to $(*)$. I cannot understand where am I making a mistake. Is (**) wrong? Thanks for any help.

Answer 1

I'm pretty sure the solution you have in (**) is the solution to (*) if you replace $\frac{d}{dt}$ with $\frac{\partial}{\partial t}$ and leave $p$ a constant throughout the problem. That is clearly not the intention. Hence, the problem is more involved.