Linear constant coefficient differential/difference equations (LCCDE)

$\displaystyle \sum_{i=0}^N a_i \frac{d^i}{dt^i}y(t)=\sum_{i=0}^N a_i y^{(i)}(t)=0$ (1)

with $N$ initial conditions $y_0,\cdots,y_{N-1}$, we assume the solution takes the form $y(t)=c e^{st}$, i.e., $y^{(i)}=c s^i e^{st}$, and convert the differential equation above into an algebraic equation:

$\displaystyle \sum_{i=0}^N a_i y^{(i)}=\sum_{i=0}^N a_i cs^ie^{st}=ce^{st}\sum_{i=0}^N a_is^i
=ce^{st} \rho(s)=0$ (2)

where $\rho(s)=\sum_{i=0}^N a_is^i$ is the characteristic polynomial. Solving the equation $\rho(z)=0$ we can find $N$ roots $s_0,\cdots,s_{N-1}$ in general, and the solution of the differential equation can be found as a linear combination:

$\displaystyle y_n=\sum_{j=0}^{N-1}c_js_j^n$ (3)

By letting $n=0,\cdots,N-1$, we can find the $N$ coefficients $c_0,\cdots,c_{N-1}$ based on the $N$ initial conditions.

To solve an Nth order LCCDE

$\displaystyle \sum_{i=0}^N a_i y_{n-i}=0$ (4)

with $N$ initial conditions $y_0,\cdots,y_{N-1}$, we assume the solution takes the form $y_n=c z^n$, i.e., $y_{n-i}=c z^{n-i}$, and convert the difference equation above into an algebraic equation:

$\displaystyle \sum_{i=0}^N a_i y_{n-i}=\sum_{i=0}^N a_i cz^{n-i}=cz^i\sum_{i=0}^N a_iz^i
=cz^i \rho(z)=0$ (5)

where $\rho(z)=\sum_{i=0}^N a_iz^i$ is the characteristic polynomial. Solving the equation $\rho(z)=0$ we can find $N$ roots $z_0,\cdots,z_{N-1}$ in general, and the solution of the difference equation can be found as a linear combination:

$\displaystyle y_n=\sum_{j=0}^{N-1}c_jz_j^n$ (6)

By letting $n=0,\cdots,N-1$, we can find the $N$ coefficients $c_0,\cdots,c_{N-1}$ based on the $N$ initial conditions.