Differential Equations: Second Order Equations

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SUMMARY

The discussion focuses on finding a second-order differential equation for which the functions y=2e^-t, y=2te^-t, and y=e^(-t+1) serve as solutions. It is established that these functions are not all independent due to the relationship between y=e^(-t+1) and y=e^-t. The characteristic equation of a linear differential equation with constant coefficients is crucial for determining independent solutions, specifically in the form (r-a)(r-b)=0.

PREREQUISITES
  • Understanding of second-order differential equations
  • Familiarity with characteristic equations
  • Knowledge of linear differential equations with constant coefficients
  • Basic calculus and exponential functions
NEXT STEPS
  • Study the method of solving second-order linear differential equations
  • Learn about the characteristic equation and its applications
  • Explore the concept of linear independence in solutions of differential equations
  • Investigate the method of undetermined coefficients for non-homogeneous equations
USEFUL FOR

Students studying differential equations, educators teaching advanced calculus, and anyone interested in the application of linear differential equations in mathematical modeling.

peace-Econ
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Homework Statement



Find a second order differential ewuation for which three functions y=2e^-t, y=2te^-t, y=e^(-t+1) are solutions.

Homework Equations





The Attempt at a Solution

 
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Since you didn't show us what you have tried, instead of answering your question directly, I will ask you this:

Can you solve this de:

y'' - 4y' + 4y = e2t

and if so, what method would you use?

Hint: This isn't an idle question.
 
peace-Econ said:

Homework Statement



Find a second order differential ewuation for which three functions y=2e^-t, y=2te^-t, y=e^(-t+1) are solutions.

Homework Equations





The Attempt at a Solution

If those functions were independent, this would be impossible but [itex]e^{-t+ 1}= e^{-t}e^1= e e^{-t}[/itex], a constant times [itex]e^{-t}[/itex] so you really have only two independent solutions.

Do you know what the "characteristic equation" of a linear differential with constant coefficients is? Such an equation will have [itex]e^{ax}[/itex] and [itex]e^{bx}[/itex] as independent solutions if and only if its characteristic equation is [itex](r- a)(r- b)= 0[/itex].
 
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