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If you find a parameter factorized in front ofįor an eigenvalue and an associated eigenvector, a straight-line solution will be given by Once an eigenvalue is found from the characteristic polynomial, then we look for the eigenvectors associated to it through the matricial equation This is a quadratic equation which has one double real root, or two distinct real roots, or two complex roots. Ĭonclusion In order to find the straight-line solution to the homogeneous linear systemįirst, we look for the eigenvalues through the characteristic polynomial Note that we have all of the eigenvectors associated to the eigenvalue. Let be an eigenvector associated to the eigenvalue. Īnswer: In the above example we checked that in fact is an eigenvalue of the given matrix. You must keep in mind that if is an eigenvector, then is also an eigenvector.įind all the eigenvectors associated to the eigenvalue. Since is known, this is now a system of two equations and two unknowns. Then, the above matricial equation reduces to the algebraic system An eigenvector associated to is given by the matricial equation These are the eigenvalues of the matrix.Īssume is an eigenvalue of the matrix A. This equation is called the Characteristic Polynomial of the system.Įxample: Find the characteristic polynomial and the eigenvalues of the matrixĪnswer: The characteristic polynomial is given by Note that the above equation is independent of the vector. Since both and can not be equal to zero at the same time, we must have the determinant of the system equal to zero. This equation may be rewritten as the algebraic system Then there must exist a non-zero vector, such that. Our next target is to find out how to search for the eigenvalues and eigenvectors of a matrix.Īnd assume that is an eigenvalue of A. Clearly, if is an eigenvector associated to, then is also an eigenvector associated to. The constant is called an eigenvalue of the matrix A, and is called an eigenvector associated to the eigenvalue of the matrix A. Where is a non-zero constant vector which satisfies Theorem: Straight-Line Solutions Consider the homogeneous linear systemĪny straight-line solution may be found in the form So, we have found two straight-line solutionsĪre these the only straight-lines? The answer is: "yes," but this will be discussed later. The first equation reduces to, or equivalently. If, then from the second equation we get. Is a straight-line solution to the system.Ĭase 2. We may ignore the constant (see the above remark). In this case, the second equation forces therefore we have If, then (since is not the zero vector). ![]() Let us illustrate the above ideas with an example.Įxample: Find any straight-line solution to the systemĪnswer: First, let us find the constant vector Note that we don't have to keep the constant C (read the above remark). Where C is an arbitrary constant, and is a non-zero constant vector which satisfies So, if a straight-line solution exists, it must be of the form Denote it byĬlearly, this is a first order differential equation which is linear as well as separable. Since and are constant vectors, we deduce that is a constant function. Remark: Note that if Y( t) is a straight-line solution, then is also a straight-line solution. ![]() So, in this case, we may think of it as an object moving along a straight line. Keep in mind that the solutions of the system may describe trajectories of moving objects. The vector is the direction vector of the line on which the solution lives. Where is a constant vector not equal to the zero vector. ![]() So it is natural to investigate whether and when an homogeneous linear system has solutions which are straight-lines.Ĭonsider the homogeneous linear system (in the matricial notation)Ī straight-line solution is a vector function of the form Lie on straight lines (can you spot them?). Remark: From the above example we notice that some solutions Let us first start with an example to illustrate the technique we will be developping.Įxample: Draw the direction field of the linear systemĪnswer: The following is the direction field: In this section we will discuss the problem of finding two linearly independent solutions for the homogeneous linear system ![]() Less IS more! Plus, an easy-to-access data port is integrated into the HS 46 C-E, allowing for simplified servicing at your local STIHL Dealer. Less frustration – its Easy2Start™ technology makes starting the trimmer nearly effortless. 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Very intense exercise: 2+ hours of elevated heart rate activity.Intense exercise: 45-120 minutes of elevated heart rate activity.Exercise: 15-30 minutes of elevated heart rate activity.Extra active (very intense exercise or physical job): BMR x 1.9.Very active (6-7 days of exercise per week): BMR x 1.725. ![]() Moderately active (3-5 days of exercise per week): BMR x 1.55.Lightly active (1-3 days of exercise per week): BMR x 1.375.Sedentary (little or no exercise): BMR x 1.2.Those who are more active typically require more calories to fuel their bodies. The number of calories a person needs each day depends on how much physical activity they engage in. For women: BMR = (10 x weight in kg) + (6.25 x height in cm) – (5 x age in years) – 161.For men: BMR = (10 x weight in kg) + (6.25 x height in cm) – (5 x age in years) + 5.Mifflin-St Jeor Equation To Calculate BMR: The Mifflin-St Jeor Equation has been shown to be more accurate than the revised Harris-Benedict Equation. It was revised in 1984 to be more accurate and was used up until 1990, when the Mifflin-St Jeor Equation was introduced. The Harris-Benedict Equation was one of the earliest equations introduced. This includes the energy that the body expends to maintain the functions of our body, such as:īasic formula for calculating BMR (Basal Metabolic Rate): The BMR accounts for approximately 60-70 per cent of the calories we burn or expend. |
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