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Solving a System Equation

Posted by Jeff on Dec 30, 2009 in Differential Equations, Linear Systems |

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. :)

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