Solving a System Equation | Engineersphere.com

Why do we need to solve system equations?

Often during a course you will need to be able to solve a system equation for its roots. These roots can be complex, distinct, or repeated. These problems usually arise when working with linear systems or differential equations. A system equation is formatted as follows:

System Equation: $Q(D)y_{0}(t) = 0$

How to solve a system equation

For example purposes, I will solve a system equation with complex roots. A system equation with complex roots as a function of $\lambda$ will appear in the following format (if it does not, you need to manipulate your equation to be in the form):

$Q(\lambda) = (\lambda - \alpha - j\beta)(\lambda - \alpha + j\beta)$

Roots: $\lambda = \alpha \pm j\beta$

So we have $y_{0}(t) = C_{1}e^{(\alpha + j\beta)t}+C_{2}e^{(\alpha - j\beta)t}$

which also equals $y_{0}(t) = Ce^{\alpha t}cos(\beta t + \theta)$

so your first step is to look at your equation and determine your roots, then write out your $y_{0}(t)$ equation with constants.

Example $\frac{d^{2}v}{dt^{2}} + 4\frac{dv}{dt} + 4v(t) = 0$ with initial conditions $V(0) = 3v$ and $V^{1}(0) = -4v$

$Q(\lambda) = \lambda^{2} + 4\lambda + 4 = 0$ $(\lambda + 2)(\lambda + 2) = 0$ $\lambda_{1} = \lambda_{2} = -2$

so now we can write our $y_{0}(t)$ equation as follows:

$y_{0}(t) = C_{1}e^{-2t}+C_{2}te^{-2t}$

In order to solve for $C_{1}$ and $C_{2}$ we need to use our initial conditions. To evaluate the first derivative initial condition, we must first take the derivative of our $y_{0}(t) = C_{1}e^{-2t}+C_{2}e^{-2t}$ that we just found.

$y_{0}^{1}(t) = -2C_{1}e^{-2t} - 2C_{2}*t*e^{-2t} + C_{2}e^{-2t}$

evaluating this equation with t = 0 and the response equal to -4v, we get this: $-4 = -2C_{1} + C_{2}$

evaluating our $y_{0}(t)$ equation with t = 0 and the response equal to 3v, we calculate $C_{1} = 3$

Using these two equations, we calculate our constants:

$C_{1} = 3$ and $C_{2} = 2$

Fill these into our $y_{0}(t)$ equation to determine the final result.

$y_{0}(t) = 3e^{-2t}+2te^{-2t}$

Now you know how to solve this common differential equations and linear systems problem, determine characteristic roots and modes, and write system equations. 🙂

Why do we need to solve system equations?

System Equation:

How to solve a system equation

Leave a Reply Cancel reply

System Equation: $Q(D)y_{0}(t) = 0$