I am starting to learn transient analysis series RC which requires our own equations to be derived, but I don't understand how the voltage across the capacitor is given in the equation below from a KVL loop around the circuit (underlined in yellow):
I'm assuming it comes from the basic capacitor voltage/current relationship formula below, but would appreciate if someone would explain how it is given in this form mathematically.
Thanks
Your questions aren't entirely clear to me, so please accept my apologies if I wander just a bit. I'll avoid using integrating factors below, as well, since it's not necessary for this problem.
For ideal capacitors (few are), the basic idea is \$Q=C\:V\$. Taking the full derivative, this is \$\text{d} Q = C\:\text{d} V+V\:\text{d}C\$. But it's usually assumed that \$\text{d}C=0\$ -- that there is no change in capacitance, since it is a fixed value in an ideal capacitor. So \$\text{d} Q = C\:\text{d} V\$. You can now introduce time by simply dividing both sides by \$\text{d} t\$, giving \$\frac{\text{d} Q}{\text{d} t} = C\:\frac{\text{d} V}{\text{d} t}=I\$. (The ad-hoc introduction of the infinitesimal of time can be done where it pleases.)
Your KVL loop equation is:
$$\begin{align*} V &= R\:I_{\left(t\right)}+\frac{1}{C}\int I_{\left(t\right)}\:\text{d}t\tag{1}\\\\ \text{d} V&=R\:\text{d}I_{\left(t\right)}+\frac{I_{\left(t\right)}}{C}\text{d} t=0\tag{2}\\\\ \text{d}I_{\left(t\right)}&=-\frac{I_{\left(t\right)}}{R\,C}\text{d} t\tag{3}\\\\ \frac{\text{d}I_{\left(t\right)}}{I_{\left(t\right)}}&=\frac{-1}{R\,C}\text{d} t\tag{4}\\\\ \int\frac{\text{d}I_{\left(t\right)}}{I_{\left(t\right)}}&=\frac{-1}{R\,C}\int\text{d} t\tag{5}\\\\ \operatorname{ln}\left(I_{\left(t\right)}\right)&=\frac{-t}{R\,C} + A_0\tag{6}\\\\ I_{\left(t\right)}&=e^{\left[\frac{-t}{R\,C} + A_0\right]}=e^{A_0}e^{\frac{-t}{R\,C}}=A\,e^{\frac{-t}{R\,C}}\tag{7} \end{align*}$$
Where \$A\$ is the constant of integration you need to worry about, now. At \$t=0\$, it must be that \$A=I_{t=0}\$. If the voltage across the capacitor at \$t=0\$ is \$V_C=0\:\text{V}\$, then it follows that \$A=\frac{V}{R}\$. So the result must be:
$$I_{\left(t\right)}=\frac{V}R\,e^{\frac{-t}{R\,C}}\tag{Series Current}\label{E1}$$
And,
$$\begin{align*} V_{C\left(t\right)} &= \frac{1}{C}\int I_{\left(t\right)}\:\text{d}t\tag{8}\\\\ &=\frac{1}{C}\int \frac{V}R\,e^{\frac{-t}{R\,C}}\:\text{d}t\tag{9}\\\\ &=\frac{V}{R\,C}\int e^{\frac{-t}{R\,C}}\:\text{d}t\tag{10}\\\\ &=\frac{V}{R\,C}\left[-R\,C\,e^{\frac{-t}{R\,C}} + A_0\right]\tag{11}\\\\ &=V\left[\frac{A_0}{R\,C}-e^{\frac{-t}{R\,C}}\right]\tag{12} \end{align*}$$
Again, at \$t=0\$ and assuming that \$V_{C\left(t=0\right)}=0\:\text{V}\$, it follows that \$A_0=R\,C\$. So:
$$V_{C\left(t\right)}=V\left[1-e^{\frac{-t}{R\,C}}\right]\label{E2}\tag{Capacitor Voltage}$$
That's it. The \$\ref{E1}\$ equation provides the current in both \$R\$ and \$C\$, since both must be the same. The \$\ref{E2}\$ equation provides the voltage that accumulates onto the capacitor over time.
Simple answer is after switch closes, Ip=V/R then decays with a slope dV/dt=V/RC = V/T then exponential decay. So Vc rises with this slope with same exponential rise toward Vc with a linear asymptote at the t=T=RC at V=~64%Vcc then slower as current reduces.
