• RC first-order system with zero input (homogeneous):

  Taking the unilateral Laplace transform of the original DE:

    $$\tau\frac{d}{dt} v_c(t)+v_c(t)=0$$ (1)
  we get
    $${\cal L}\left[\tau\frac{d}{dt} v_c(t)+v_c(t)\right] =\tau [s V_c(s)-V_0]+V_c(s)=V_c(s)(s\tau +1)-\tau V_0=0$$ (2)
  Here we have used the property ${\cal L}(dx/dt)=s X(s)-x(0)$, where $x(0)=\lim_{t\rightarrow 0}x(t)$ is the initial condition. Solving for $V_c(s)$ we get
    $$V_c(s)=\frac{\tau V_0}{s\tau+1}=\frac{V_0}{s+1/\tau}$$ (3)
  The time domain voltage is the inverse transform:
    $$v_c(t)={\cal L}^{-1}\left[\frac{V_0}{s+1/\tau}\right]=V_0 e^{-t/\tau}$$ (4)

• RC first-order system with constant input:

  Taking the unilateral Laplace transform of the original DE:

    $$\tau\frac{d}{dt} v_c(t)+v_c(t)=v_s(t)=V_s\,u(t)$$ (5)
  we get
    $${\cal L}\left[\tau\frac{d}{dt} v_c(t)+v_c(t)\right] =\tau [s V_c(s)-V_0] + V_c(s)={\cal L} \left[ V_s\,u(t) \right]=\frac{V_s}{s}$$ (6)
  Solving for $V_c(s)$, we get
    $$V_c(s)=\frac{V_s}{s(s\tau+1)}+\frac{\tau V_0}{s\tau+1} =V_s\left(\frac{1}{s}-\frac{\tau}{s\tau+1}\right)+\frac{\tau V_0}{s\tau+1}$$ (7)
  Taking the inverse Laplace trannsform we get the solution of the DE in time domain:
    $$v_c(t)={\cal L}^{-1}\left[V_c(s)\right]=V_s-V_s e^{-t/\tau}+V_0e^{-t/\tau} =V_s+(V_0-V_s)e^{-t/\tau}$$ (8)
  To find the steady state solution, we can use the final-value theorem:
    $$\lim_{t\rightarrow\infty}v_c(t)=\lim_{s\rightarrow 0} s V_c(s) =... ...rightarrow 0} \left[\frac{V_s}{(s\tau+1)}+\frac{\tau sV_0}{s\tau+1}\right]=V_s$$ (9)

• RC first-order system with sinusoidal input:

  First consider a complex input $V_s e^{j\omega t}\,u(t)=V_s(\cos\omega t+j\sin\omega t)u(t)$ on the right-hand side:

    $$\tau\frac{d}{dt} v_c(t)+v_c(t)=V_s \cos(\omega t)=V_s e^{j\omega t}\,u(t)$$ (10)
  Taking the transform on both sides, we get
    $${\cal L}\left[\tau\frac{d}{dt} v_c(t)+v_c(t)\right] =s\tau V_c(s... ...u V_0+V_c(s)={\cal L}\left[V_s\,e^{j\omega t}u(t)\right]=\frac{V_s}{s-j\omega}$$ (11)
  Solving for $V_c(s)$, we get
    $$V_c(s)=\frac{1}{s\tau+1}\left(\frac{V_s}{s-j\omega}+\tau V_0\righ... ...\left(\frac{1}{s-j\omega}-\frac{1}{s+1/\tau}\right) +\frac{\tau V_0}{s\tau+1}$$ (12)
  Taking the inverse Laplace trannsform we get the solution of the DE in time domain:
  

  $$v_c(t) = {\cal L}^{-1}\left[V_c(s)\right] =\frac{V_s}{j\omega\tau+1}\left[... ...1}{s+1/\tau}\right)\right]+{\cal L}^{-1}\left( \frac{\tau V_0}{s\tau+1} \right)$$

  $$= \frac{V_se^{-j\phi}}{\sqrt{\omega^2\tau^2+1}}\left[e^{j\omega t}-... ...phi)} -\frac{V_se^{-j\phi}}{\sqrt{\omega^2\tau^2+1}}e^{-t/\tau}+V_0 e^{-t/\tau}$$

  
  
  Taking the real part we get the solution:
  

  $$v_c(t) = \frac{V_s}{\sqrt{\omega^2\tau^2+1}}\cos(\omega t-\phi) -\frac{V_se^{-t/\tau}}{\sqrt{\omega^2\tau^2+1}}\cos(-\phi) +V_0 e^{-t/\tau}$$

  $$= \frac{V_s}{\sqrt{\omega^2\tau^2+1}}\cos(\omega t-\phi) +\left[V_0 -\frac{V_s}{\sqrt{\omega^2\tau^2+1}}\cos(-\phi)\right] e^{-t/\tau}$$