Linear Constant Coefficient ODE (LCCODE)

We first consider the complete solution of a first order LCCODE:

$$\frac{d}{dt} y(t)+ay(t)=\dot{y}(t)+ay(t)=x(t)$$

is $y(t)=y_h(t)+y_p(t)$, the sum of the homogeneous solution $y_h(t)$ and a particular solution $y_p(t)$.

• To find $y_h(t)$ that satisfies the homogeneous DE $\dot{y}(t)+ay(t)=0$, we assume $y_h(t)=c\;e^{st}$, and substitute it together with $\dot{y}(t)=cs\;e^{st}$ into the DE and get
  
  $$\dot{y}(t)=cs e^{st}+ac e^{st}=0$$
  
  
  Dividing both sides by $c e^{st}\ne 0$, we get $s=-a$, i.e, $y_h(t)=c e^{-at}$. To determine the coefficient $c$, we evaluate $y_h(t)$ at $t=0$ and get
  
  $$y_h(t)\bigg\vert _{t=0}=c e^{-at}\bigg\vert _{t=0}=c=y_0$$
  
  
  where $y_0=y(0)$ is the initial condition assumed to be given. Then we have
  
  $$y_h(t)=e^{-at} y_0$$
  
  

• To find the complete solution, we multiply both sides of the ODE by $e^{at}$ and get
  
  $$e^{at} \left[ \dot{y}(t)+a y(t) \right] =e^{at} \dot{y}(t)+a e^{at}y(t)=e^{at} x(t)$$
  
  
  which can be rewritten as
  
  $$\frac{d}{dt} \left[ e^{at}y(t) \right]=e^{at}x(t)$$
  
  
  Integrating both sides we get
  
  $$\int_0^t \left[\frac{d}{dt} \left( e^{a\tau}y(\tau) \right)... ...igg\vert _0^t=e^{at}y(t)-y(0)=\int_0^t e^{a\tau}x(\tau) d\tau$$
  
  
  Multiplying both sides by $e^{-at}$ and rearranging, we get the expression for $y(t)$
  
  $$y(t)=e^{-at}\left[ y(0)+\int_0^t e^{a\tau}x(\tau) d\tau \ri... ...-at}y_0+\int_0^t e^{-a(t-\tau)}x(\tau) d\tau =y_h(t)+y_p(t)$$
  
  
  where $y_h(t)=e^{-at} y_0$ is the homogeneous solution found above, and
  
  $$y_p(t)=\int_0^t e^{-a(t-\tau)}x(\tau) d\tau=e^{-at}\;*\;x(t)$$
  
  
  In particular, if the input $x(t)=\delta(t)$ is an impulse, we get the output $y(t)$ as the impulse response to be $y_p(t)=e^{-at}$. We therefore see that in general, the particular solution is the convolution of the impulse response $e^{-at}$ and the input $x(t)$.

The DE can also be solved in s-domain based on Laplace transform. Taking Laplace transform on both sides of the DE, we get

$${\cal L}\left[ \;\dot{y}(t)+ay(t)\;\right] =sY(s)-y(0)={\cal L}\left[ \;x(t)\;\right]=X(s)$$

Solving for $Y(s)$ we get

$$Y(s)=\frac{y_0}{s+a}+\frac{X(s)}{s+a}$$

Taking inverse transform we get

$$y(t)={\cal L}^{-1} \left[\frac{y_0}{s+a}\right] +{\cal L}^{-1}\left[ \frac{X(s)}{s+a}\right] =y_0e^{-at}+x(t)\;*\;e^{-at}$$

We next consider the complete solution of a second order LCCODE in the following canonical form:

$$\ddot{y}(t)+2\zeta\omega_n \dot{y}(t)+\omega_n^2 y(t)=x(t)$$

Same as in the first order case, the complete solution of this second order LCCODE is composed of the homogeneous solution $y_h(t)$ when $x(t)=0$, and the particular solution $y_p(t)$ when $x(t)\ne 0$.