How Do You Calculate Specific Values for Constants in Differential Equations?

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Homework Help Overview

The discussion revolves around solving a system of differential equations and determining specific values for constants based on initial conditions. The equations involve variables x, y, and z, with the original poster seeking guidance on how to find the constants after deriving the general solution.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss substituting t=0 into the general solutions to find specific values for the constants. There is a focus on setting the resulting expressions equal to the given initial conditions (x,y,z)=(2,3,-1) and questioning how to express the solutions in terms of the constants.

Discussion Status

Some participants have provided guidance on how to approach the problem, including suggestions for simplifying the equations by substituting t=0. Multiple interpretations of how to express the solutions and the constants are being explored, but there is no explicit consensus on the final steps to take.

Contextual Notes

Participants note the lack of explicit instructions from the professor regarding the method to find the constants, leading to some uncertainty in the discussion. The original poster is working within the constraints of a homework assignment that requires specific values based on initial conditions.

hbomb
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Give the general solution of the set of equations below:

x'=5x-2y+3z
y'=y
z'=6x+7y-2z

Which I found to be:
x=3C1e^7t-3C2e^t-C3e^-4t
y=6C2e^t
z=2C1e^7+8C2e^t+3C3e^-4t

Here's where I'm stuck. They want me to find the solution at t=0, (x,y,z)=(2,3,-1)

Which the professor hasn't told us how to do this.
 
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I'm sure you know how to do this. If t=0, what is x(0)? There are two ways to express this... one is from the equation for x you solved for, the other is from the point you were given. Since t=0, all those nasty exponentials go away, and you have a system of three variables with constant coefficients (which is remarkably simpler than the initial set of differential equations you solved)
 
ok, so what I understand from this is that I plug t=0 into all the t's of the general solutions.

x=3C1-32-C3
y=6C2
z=2C1+8C2+3C3

And then do I set these equal to (x,y,z)=(2,3,-1). What is the solutions form suppose to look like?
 
Let's call [tex]C_1 C_2[/tex] and [tex]C_3[/tex] A, B, and C to make it easier.

You know
x(0)=3A - 3B - C=2
y(0)=6B = 3
z(0)=2A + 8B + 3C=-1

So you should be able to solve for A, B and C as numbers. For example, if 6B=3, B=2. Now you know 3A-6-C = 2, and 2A + 16 + 3C = -1
 

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