Solve 4th Order Linear ODE and Plot Graph: t→∞

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    Linear Ode
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Discussion Overview

The discussion revolves around solving a fourth-order linear ordinary differential equation (ODE) and analyzing the behavior of its solution as \(t\) approaches infinity. Participants explore the characteristic equation, propose potential solutions, and discuss the implications of repeated roots.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents the ODE \(y^{(4)}-4y'''+4y''=0\) and attempts to derive the characteristic equation, leading to roots \(r=0\) and \(r=2\).
  • Another participant suggests a solution form \(y=c_1+c_2e^{2t}+c_3te^{2t}\) but expresses uncertainty about its correctness.
  • A later reply introduces a substitution \(y^{\ ''}= z\) to transform the ODE, leading to a new equation \(z^{\ ''} - 4z^{\ '} + 4z = 0\) and proposes a solution for \(z(t)\) involving exponential terms.
  • Another participant corrects the proposed solution form, indicating that the general solution should account for repeated roots, suggesting \(y(t)=c_1+c_2t+c_3e^{2t}+c_4te^{2t}\).

Areas of Agreement / Disagreement

Participants express uncertainty about the correctness of their proposed solutions and interpretations of the characteristic roots. There is no consensus on the final form of the solution, and multiple competing views remain regarding the correct approach to solving the ODE.

Contextual Notes

Participants have not fully resolved the implications of the repeated roots on the general solution form, and there are unresolved mathematical steps in the integration process.

alane1994
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Find the solution of the given initial value problem, and plot its graph. How does the solution behave as \(t\rightarrow\infty\)

\(y^{(4)}-4y'''+4y''=0\)

My work, which coincidentally I believe is incorrect...

From the above differential equation,

\(r^4-4r^3+4r^2=0\)

\(r^2(r-2)^2=0\)

\(r=0,~2\)

Then, would I just input that into a form like this?

\(y=c_1+c_2e^{2t}+c_3te^{2t}\)I definitely feel as though this isn't correct...
 
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alane1994 said:
Find the solution of the given initial value problem, and plot its graph. How does the solution behave as \(t\rightarrow\infty\)

\(y^{(4)}-4y'''+4y''=0\)

My work, which coincidentally I believe is incorrect...

From the above differential equation,

\(r^4-4r^3+4r^2=0\)

\(r^2(r-2)^2=0\)

\(r=0,~2\)

Then, would I just input that into a form like this?

\(y=c_1+c_2e^{2t}+c_3te^{2t}\)I definitely feel as though this isn't correct...

Setting $\displaystyle y^{\ ''}= z$ the ODE becomes...

$\displaystyle z^{\ ''} - 4\ z^{\ '} + 4\ z\ (1)$

... the solution of which is...

$\displaystyle z(t)= c_{1}\ e^{2\ t} + c_{2}\ t\ e^{2\ t}\ (20)$

Now You can solve the ODE...

$\displaystyle y^{ ''} = z\ (3)$

... with two successive integration obtaining... $\displaystyle y^{\ '} = \int z(t)\ dt\ (4)$ $\displaystyle y = \int y^{ '} (t)\ dt\ (5)$ Kind regards $\chi$ $\sigma$
 
Thank you very much! That makes quite a bit more sense!
 
alane1994 said:
Find the solution of the given initial value problem, and plot its graph. How does the solution behave as \(t\rightarrow\infty\)

\(y^{(4)}-4y'''+4y''=0\)

My work, which coincidentally I believe is incorrect...

From the above differential equation,

\(r^4-4r^3+4r^2=0\)

\(r^2(r-2)^2=0\)

\(r=0,~2\)

Then, would I just input that into a form like this?

\(y=c_1+c_2e^{2t}+c_3te^{2t}\)I definitely feel as though this isn't correct...

Both of your characteristic roots are repeated, so the general form of your solution would be:

$$y(t)=c_1+c_2t+c_3e^{2t}+c_4te^{2t}$$
 

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