Second order linear system and power series: Differential Equations

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SUMMARY

The discussion focuses on finding a third-degree polynomial approximation for the general solution to the differential equation \(\frac{d^{2}y}{dt^{2}} + 3\frac{dy}{dt} + 2y = \ln(t+1)\). The homogeneous solution is established as \(y(t) = k_1 e^{-t} + k_2 e^{-2t}\). A proposed solution of the form \(\frac{at^{3}}{3} - \frac{bt^{2}}{2} + ct\) leads to coefficients \(a = \frac{1}{2}\), \(b = 2\), and \(c = \frac{2}{3}\). The discussion concludes that while a third-degree polynomial can approximate the solution, it cannot fully resolve higher-order terms such as \(t^{4}\) and \(t^{5}\).

PREREQUISITES
  • Understanding of differential equations, specifically second-order linear systems.
  • Familiarity with power series expansions, particularly for \(\ln(t+1)\).
  • Knowledge of homogeneous solutions and their applications in differential equations.
  • Basic skills in polynomial approximation techniques.
NEXT STEPS
  • Study the method of undetermined coefficients for non-homogeneous differential equations.
  • Explore power series solutions for differential equations in more depth.
  • Learn about the Laplace transform and its application in solving linear differential equations.
  • Investigate the concept of asymptotic analysis for approximating solutions to differential equations.
USEFUL FOR

Students and educators in mathematics, particularly those focusing on differential equations, as well as engineers and physicists who require polynomial approximations in their analyses.

clarineterr
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Homework Statement


Find a third degree polynomial approximation for the general solution to the differential equation:

[tex]\frac{d^{2}y}{dt^{2}}[/tex] +3[tex]\frac{dy}{dt}[/tex]+2y= ln(t+1)

Homework Equations


Power series expansion for ln(t+1)


The Attempt at a Solution



The system to the corresponding homogeneous equation [tex]\frac{d^{2}y}{dt^{2}}[/tex] +3[tex]\frac{dy}{dt}[/tex]+2y = 0

is y(t) = k1e-t +k2e-2t

Then I guessed[tex]\frac{ at^{3}}{3}[/tex]-[tex]\frac{bt^{2}}{2}[/tex]+ct as a solution for the original equation. Plugging this in I got a=1/2, b=2,c=2/3

But then I still have the t[tex]^{4}[/tex], t[tex]^{5}[/tex] terms, etc left in the equation. I am not quite sure how a third degree polynomial can be a solution to this equation.
 
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clarineterr said:
I am not quite sure how a third degree polynomial can be a solution to this equation.
Because it will be an approximation not really the solution itself.
 

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