Introductuin to differential equations

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Homework Help Overview

The discussion revolves around identifying constant solutions for a specified differential equation, specifically y'' + 4y' + 6y = 10. Participants are exploring the implications of setting derivatives to zero to determine the nature of solutions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to integrate the second derivative and substitute results back into the equation. Some participants question the necessity of integration, suggesting a more direct approach by setting derivatives to zero. Others explore the implications of the derived expressions for y.

Discussion Status

There is an ongoing exploration of the problem, with participants offering guidance on simplifying the approach. While some interpretations are being discussed, there is no explicit consensus on the correctness of the proposed solutions.

Contextual Notes

Participants are navigating the constraints of the problem, particularly regarding the assumptions about constant solutions and the validity of y = 0 as a potential solution.

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Homework Statement


use the concept that y = c, -inf < x < inf is a constant function if and only if y' = 0 to determine whether the specified differential equation has any constant solutions:

y'' + 4y' + 6y = 10

The Attempt at a Solution



What throws me off in this particular problem is y''. If y' = 0 then y'' = 0 as well.
Only now when I integrate y'' to y' its y' = c1 so c1 = 0 and y' still = 0.

then I basically substitute my results in the equation to get:

(0) + 4(0) + 6(c) = 10

6c = 10

[c = 10/6, c1 = 0] are the constant solutions to the equation.

Does this make sense?

Thanks.
 
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You are thinking too hard. You don't have to integrate anything. Set y''=y'=0 and solve the equation for y. y=0 is NOT a solution.
 
ahh.. so

y = (10 - 4y' - y'')/6

y = (10 - 4(0) - 0)/6

[y = 10/6] ?

Thanks for the help! Please let me know if this is right.
 
The great thing about ODE's is that you can usually quite readily check the answer by substituting your solution back into the differential equation. Give it a try and you'll find that your solution is correct.
 

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