Solving the Fourth-Order Differential Equation: d4y/dt4 - λ4 y= 0

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Discussion Overview

The discussion revolves around solving the fourth-order differential equation d4y/dt4 - λ4 y= 0. Participants explore methods for finding solutions, including the characteristic equation and its roots, while addressing varying levels of understanding among contributors.

Discussion Character

  • Homework-related
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant asks how to solve the fourth-order differential equation.
  • Another participant provides the characteristic equation x^4 - λ^4 = 0 and factors it, identifying the characteristic values as λ, -λ, iλ, and -iλ.
  • A participant expresses confusion about the explanation and requests a clearer working, questioning the notation used.
  • Further inquiry is made about solving the equation x^2 + λ^2, leading to a clarification that x^2 + λ^2 is not an equation, but x^2 + λ^2 = 0 was solved by taking the square root of both sides.
  • There is a challenge regarding the participant's understanding of basic quadratic equations in the context of solving differential equations.

Areas of Agreement / Disagreement

The discussion reflects a lack of consensus, with varying levels of understanding and differing approaches to the problem. Some participants are focused on the technical aspects of the solution, while others express confusion and seek clarification.

Contextual Notes

Participants' responses indicate varying assumptions about prior knowledge of linear differential equations and quadratic equations, which may affect the clarity of the discussion.

teeoffpoint
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how can i solve d4y/dt4 - λ4 y= 0
 
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Since that is a linear equation with constant coefficients, you can immediately write down its "characteristic equation", [itex]x^4- \lambxa^4= 0[/itex]. That factors as [itex](x^2- \lambda^2)(x^2+ \lambda^2)= (x- \lambda)(x+ \lambda)(x- i\lambda)(x+ i\lambda)= 0[/itex] and so has characteristic values [itex]\lambda[/itex], [itex]-\lambda[/itex], [itex]i\lambda[/itex], and [itex]-i\lambda[/itex]. Do you know what to do with those?
 
hmm... i don't really understand. can u show me a clearer working? wat you mean by x^4- \lambxa^4= 0? thanks
 
Last edited:
What kind of explanation would you understand? Do you know anything at all about linear differential equations? I need to know what you do understand before I can explain much more.
 
oh how did you solve x^2 + λ^2?
 
I didn't solve [itex]x^2+ \lamba^2[/itex]. That is not an equation. I did solve [itex]x^2+ \lambda^2= 0[/itex] by the obvious method: I subtracted [itex]\lambda^2[/itex] from both sides to get [itex]x^2= -\lambda^2[/itex] and then took the square root of both sides. But now, in addition to my previous questions, which you still haven't answered, why are you trying to do differential equations if you don't know how to solve a simple quadratic equation?
 

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