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Solve the linear system of ODE ##x' = 2x + 3y##, ##y' = -3x + y## with initial conditions ##x(0) = 1, y(0) = 2##.
julian said:By the way, there are other methods of solving the problem!
Yes, they are functions of ##t##.Mayhem said:When working with these types of problems is it implied that x = x(t) and y = y(t).
Because you have a second order differential equation the general solution will involve two unknown constants, so you need ##x(0)## and ##x'(0)##. Now, ##x'(0)## was not specified in the question but that does not present a problem because you are given a formula for ##x'(t)##. Yep?erobz said:The system is equivalent to two second order ODE's
$$ y'' - 3y'+11y , y(0)=2$$
$$ x''-3x'+11x=0, x(0)=1 $$
Algebraic error fixed ( hopefully)...
Using the original equations and given initial conditions we find ## x'(0) = 8, y'(0) = -1##julian said:Because you have a second order differential equation the general solution will involve two unknown constants, so you need ##x(0)## and ##x'(0)##. Now, ##x'(0)## was not specified in the question but that does not present a problem because you are given a formula for ##x'(t)##. Yep?
julian said:Do you know how to go about solving your differential equation? It is a homogeneous linear differential equation of second order with constant coefficients.
But different constants in each case, not the same.erobz said:Using the original equations and given initial conditions we find ## x'(0) = 8, y'(0) = -1##
With an old textbook in front of me... Find the auxiliary equation:
$$m^2 -3m+11 = 0 \implies m = \frac{3}{2} \pm \frac{\sqrt{35}i}{2} $$
I think that implies the general solution ( for ##y## or ##x## - same ODE - different i.c. ):
$$ x(t) = y(t) = e^{ \left( \frac{3}{2}t \right) } \left( c_1 \cos \left( \frac{ \sqrt{35} }{2} t \right) + c_2 \sin \left( \frac{ \sqrt{35} }{2} t \right) \right) $$
Yeah, I was just being lazy. Thats why I said the general solution is... the constants would be determined for the initial conditions of each. I would think those constants aren't necessarily distinct in the general solution...so I thought I could get away with it?bob012345 said:But different constants in each case, not the same.
A linear system of ODEs (ordinary differential equations) is a set of equations that describe the relationship between multiple variables and their derivatives. The equations are linear, meaning that the variables and their derivatives are raised to the first power and there are no products or powers of the variables. An example of a linear system of ODEs is x' = 2x + y and y' = -3x + 4y.
To solve a linear system of ODEs, you can use a variety of methods such as substitution, elimination, or matrix operations. The specific method used will depend on the complexity of the system and the desired form of the solution. In general, the goal is to isolate the variables and their derivatives on one side of the equations and the constants on the other side.
Initial conditions refer to the values of the variables at a specific starting point, usually denoted as t = 0. In the context of solving a linear system of ODEs, initial conditions are used to find the specific solution to the system. These initial conditions can be given as specific values or as a set of equations.
Initial conditions are important because they help determine the specific solution to a linear system of ODEs. Without initial conditions, the solution would be a general solution that applies to all possible values of the variables. However, with initial conditions, the solution can be tailored to fit a specific scenario or problem.
Linear systems of ODEs have many real-world applications, including modeling population growth, chemical reactions, and electrical circuits. They are also used in engineering and physics to model and predict the behavior of systems and processes. In economics, linear systems of ODEs can be used to analyze supply and demand, investment growth, and other economic trends.