Engineer Sphere » Differential Equations http://engineersphere.com All the cool kids are doing it. Thu, 08 Jul 2010 03:23:29 +0000 en hourly 1 http://wordpress.org/?v=3.0 Solving a System Equation http://engineersphere.com/math/differential-equations/solving-a-system-equation.html http://engineersphere.com/math/differential-equations/solving-a-system-equation.html#comments Wed, 30 Dec 2009 00:02:11 +0000 Jeff http://engineersphere.com/?p=1117 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

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. :)

]]> http://engineersphere.com/math/differential-equations/solving-a-system-equation.html/feed 0 Laplace Transforms http://engineersphere.com/math/laplace-transforms.html http://engineersphere.com/math/laplace-transforms.html#comments Sun, 06 Sep 2009 18:06:20 +0000 Luke http://engineersphere.com/?p=701 The Laplace Transform is a method that simplifies integral and differential equations into algebraic equations. This practice is commonly used to solve for a function out of a differential equation, which otherwise may have been unsolvable or very difficult. The following integrals can be used to transform between F(s) \Leftrightarrow f(t) where \mathcal{L} denotes Laplace and \mathcal{L}^{-1} denotes Inverse Laplace:

\mathcal{L} [f(t)] = F(s) = \int\limits_0^\infty {f(t)*e ^ {-st}d} t

\mathcal{L}^{-1}[F(s)] = f(t) = \frac{1}{2 \pi j} \int\limits_{c - j \omega}^{c + j \omega} {F(s)*e ^ {-st}d}s

Since these integrals can be tedious and certain functions tend to reoccur, a table of Laplace Transforms has been linked:

Laplace Transforms Table

This table can be a little complex to use at first so an example is provided below to get you started. In this problem we implement the Laplace Transform and Inverse Laplace Transform to solve for y(t) .

\frac{d^2y}{dt^2} + 7 \frac{dy}{dt} + 12y = 10 \quad y(0)=3, y'(0)

The first step is to take the Laplace Transform of both sides of the equation. Use element 1 of our table for the right side and element 18 for the left side. Note that the initial conditions are necessary to take the Laplace Transform of the left side.

s^2 Y(s) - s y(0) - y'(0) + 7 (s Y(s) - y(0)) + 12 Y(s) = \frac{10}{s}

Inputting our initial conditions:

s^2 Y(s) - 3s - 0 + 7 (s Y(s) - 3) + 12 Y(s) = \frac{10}{s}

Assuming you are an engineering student and can do a little alegebra, our next step is to find the terms that have Y(s) in common and factor it out. Our goal is to find Y(s) = F(s) because, after all, \mathcal{L}^{-1}[Y(s)] = y(t) :

(s^2 +7s + 12)Y(s) - 3s -21 = \frac{10}{s}

After solving for Y(s) and factoring the denominator:

Y(s) = \frac{3s^2 + 21s + 10}{s(s+4)(s+3)}

Now we arrive at the trickier part of this procedure. We must take the Inverse Laplace of Y(s) to find y(t) . If our function Y(s) does not match anything in the table, such as this case, factoring is a good place to start. This problem can easily be factored using the expand function on your TI-89. Just go to catalog \rightarrow expand and enter your function in parenthesis. Using this function:

Y(s) = \frac{\frac{5}{6}}{s} - \frac{\frac{13}{2}}{s+4} + \frac{\frac{26}{3}}{s+3}

Noting \mathcal{L} [f_1(t) + f_2(t)] = \mathcal{L} [f_1(t)] + \mathcal{L} [f_2(t)] and \mathcal{L} [k*f(t)] = k*\mathcal{L} [f(t)] , we can take the Laplace Transform of each term independently and also manipulate the constant terms if necessary or just pull them out. Using the 2nd property in our Laplace Transform table:

y(t) = \frac{5}{6} - \frac{13}{2}e^{-4t} + \frac{26}{3}e^{-3t}

To check your work you can plug y(t), \frac{dy}{dt}, \frac{d^2y}{dt^2} into the original, differential equation and at t=0 we find that 10=10.

]]> http://engineersphere.com/math/laplace-transforms.html/feed 1 Separable Differential Equations http://engineersphere.com/math/separable-differential-equations.html http://engineersphere.com/math/separable-differential-equations.html#comments Mon, 27 Jul 2009 01:51:08 +0000 Jeff http://engineersphere.com/?p=447 These are 1st order differential equations, they can be written in the form y^I = f(t)g(y)

Why is it called a first order equation?

You can tell the order of any equation by looking at the highest power of the differential (\frac{\text{d}}{\text{d}x}) on either side of the equation,  the below case, we have a FIRST order differential equation because (\frac{\text{d}}{\text{d}x}) is raised to the power of 1.

\frac{\text{d}y}{\text{d}x} - y^2 e^{x} = \cos(x) + e^{x} + \cos(x) y^2

Our goal is to solve for y.  But how can we do that if we have a differential in our equation?  In order to start, we need to isolate the \frac{\text{d}y}{\text{d}x} .

\frac{\text{d}y}{\text{d}x} = \cos(x) + e^{x} + \cos(x) y^2 + y^2 e^{x}

Combine any like-terms that you can, this will make things way easier and you will struggle to finish one of these without doing this.  As you can see below, I have found e^{x} occurring more than once on the right side, so I can pull that out as a common factor.

\frac{\text{d}y}{\text{d}x} = \cos(x) + \cos(x) y^2 + e^{x} (1 + y^2)

We are not done yet, I see one more, \cos(x)

\frac{\text{d}y}{\text{d}x} = \cos(x)(1 + y^2) + e^{x} (1 + y^2)

Things are starting to look a little bit neater, but we can take this 1 step further, (1 + y^2) is a like-term as well, we now have this.

\frac{\text{d}y}{\text{d}x} = (1 + y^2)(\cos(x) + e^{x})

Note: y^{2} + 1 = 0 has no real solutions (square root of a negative number is imaginary)

Alright, what should we do from here?

We need to get all of the ‘x’ terms together and all of the ‘y’ terms together, this includes the dy and dx on the left side.  As you can see, \frac{\text{d}y}{\text{d}x} is simply a fraction, multiply the dx to the right side of the equation to end up with this.

dy = (1 + y^2)(\cos(x) + e^{x})dx Almost done, now we move all terms with ‘y’ in them to the left side like so,

\frac{\text{d}y}{(1 + y^2)} = (\cos(x) + e^{x})dx

Now we have something we are familiar with!  Using integration tables, TI-89, or just plain skills, you should now integrate each side of the equation so that you can have these derivatives (dy and dx) as simply variables (x & y).  This is simple calculus.

\int_{}^{} \frac{1}{(1 + y^2)}(dy) = \int_{}^{} \cos(x) + e^{x}(dx)

Which equates to

\arctan(y) = \sin(x) + e^{x} + C

y = \tan(\sin(x) + e^{x} + C)

What is “C”?!

The “C” represents a constant that is generated via the integration process.  When you take a derivative of a constant, the result is Zero, this must be accounted for when working backwards with an integral.  This is calculus as well.  In later lessons, we will be manipulating this constant and solving for it, in order to do this, we need initial conditions given to us (or some professors will try to be tricky by making you solve for those as well).  This constant is often represented with a “K”, as well.

We have succeeded in finding y as a function of x, we started with a very cryptic equation that included differentials, y’s and x’s jumbled up together.  As you can see, through a series of combining like-terms,  variable manipulation and integration, one can solve for an unknown variable.

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