Science & engineering math: system of differential equations

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Homework Help Overview

The discussion revolves around a system of differential equations involving two functions, y(t) and z(t), with specific initial conditions. The original poster expresses uncertainty about how to begin solving the equations and mentions a previous class example that utilized Kramer's rule.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore various approaches, including the suggestion to add the equations to simplify the system. There is also a discussion about the implications of the initial conditions and whether the equations were stated correctly, particularly regarding the presence of derivatives of z.

Discussion Status

The conversation is ongoing, with participants providing insights and questioning the setup of the problem. A clarification about the equations has been made, indicating a potential path forward by integrating the corrected first equation. However, the implications of the initial conditions remain a point of contention.

Contextual Notes

There is a noted confusion regarding the original equations, specifically the absence of a derivative for z in the initial formulation, which raises questions about the validity of the initial conditions provided.

chatterbug219
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Homework Statement



Solve the system of differential equations:
y'(t) + z(t) = t
y"(t) - z(t) = e-t
Subject to y(0) = 3, y'(0) = -2, and z(0) = 0

Homework Equations



My professor did an example in class that was much simpler and solved it using Kramer's rule.

The Attempt at a Solution


I don't know how to start it. I thought about rearranging the equations so that one was equal to y'(t) and the other was equal to z(t), but I'm not sure that would work...
 
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What about adding the equations so as z(t) cancels?

ehild
 
I would definitely start off as ehild has suggested.
 
There is, however, a problem with the entire exercise. Doing as ehild suggests gives you a second order differential equation in y only which you can solve and then use the initial conditions to give a specific solution for y. But there is no derivative of z in these equations- once you know y, z is fixed and you have no constant to choose to make z(0)= 0. Was one or both of those "z"s supposed to be z'? If not then anyone of the three conditions, y(0)= 3, y'(0)= -2, z(0)=n 0, can be dropped to give a solution but there is not y, z, satisfying the equations and all three of the conditions.
 
Oh my gosh yes there was supposed to be a z' in the first equation
So it is: y'(t) + z'(t) = t
The second equation is correct though, so sorry for any confusion!
 
chatterbug219 said:
Oh my gosh yes there was supposed to be a z' in the first equation
So it is: y'(t) + z'(t) = t
The second equation is correct though, so sorry for any confusion!

Then you can integrate the first equation and add to the second.

ehild
 

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