A differential equation question

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Discussion Overview

The discussion revolves around a differential equation, specifically y' + y = 1, and its initial condition y(0) = 2.5. Participants explore the implications of the general solution y = ce^(-x) + 1 and how to determine the constant c based on the initial condition.

Discussion Character

  • Technical explanation, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant notes that the general solution y = ce^(-x) + 1 satisfies the differential equation for all values of c, but is puzzled by the infinite number of possible c values.
  • Another participant calculates c by substituting the initial condition into the general solution, concluding that c = 1.5.
  • A later reply reiterates the calculation of c = 1.5, confirming the previous statement.
  • Another participant emphasizes that the task is to find the specific value of c that meets the initial condition y(0) = 2.5.

Areas of Agreement / Disagreement

Participants generally agree on the method to find c but express uncertainty regarding the implications of the general solution and the role of the constant c in satisfying the initial condition.

Contextual Notes

The discussion does not resolve the implications of having multiple values for c in relation to the differential equation and its initial condition.

Who May Find This Useful

This discussion may be useful for individuals studying differential equations, particularly those interested in understanding the role of initial conditions in determining specific solutions.

mech-eng
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There is differential equation with initial condition perplexing me.

y'+ y = 1, y = ce^-x + 1 , y = 2.5 when x = 0

First I take derivative of y which is -ce^-x then I sum it up with y which is -ce^-x+ce^-x + 1 equals 1 which is in harmony with y' + y = 1 but it
seems that this is independent from integral constant c and so there are infinite number of c but in the answer c is -1.5.

Have a nice day.
 
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y=1+c*e^(-x); at x= 0, y = 1+c*e° = 1+c=2.5, ...so, c = 1.5
 
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abitslow said:
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y=1+c*e^(-x); at x= 0, y = 1+c*e° = 1+c=2.5, ...so, c = 1.5

Thank you.
 
You were told, when they said "y= ce^{-x}+ 1" that this would satisfy the differential equation for all values of c. That is what that means. The problem was to find c such that y(0)= 2.5.
 
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