Finding a linear differential equation

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SUMMARY

The discussion focuses on solving a linear differential equation, specifically identifying the general solution as the sum of the null solution (complementary solution) and the particular solution. The correct formulation of the general solution is y(t) = Ae2t + 5e8t - 5, where A is a constant. The derivative of the function is calculated as y' = 2e2t + 40e8t. Participants emphasize the importance of finding the nonhomogeneous differential equation corresponding to the given solution.

PREREQUISITES
  • Understanding of linear differential equations
  • Familiarity with complementary and particular solutions
  • Knowledge of differentiation techniques
  • Experience with nonhomogeneous differential equations
NEXT STEPS
  • Study the method of undetermined coefficients for finding particular solutions
  • Learn about the theory of linear differential equations
  • Review examples of nonhomogeneous differential equations in textbooks
  • Practice solving linear differential equations using software tools like MATLAB or Mathematica
USEFUL FOR

Students studying differential equations, educators teaching calculus or differential equations, and anyone looking to deepen their understanding of linear differential equations and their solutions.

goonking
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Homework Statement


upload_2016-6-12_19-3-31.png


Homework Equations

The Attempt at a Solution


So the general solution is the sum of the null solution (Yn) and particular solution (Yp)

I believe I just need to write:

y = e2t + 5e8t - 5 + C

and then find the derivative of both sides

y' = 2e2t + 40e8t

is this correct?
 
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goonking said:

Homework Statement


View attachment 101985

Homework Equations

The Attempt at a Solution


So the general solution is the sum of the null solution (Yn) and particular solution (Yp)

I believe I just need to write:

y = e2t + 5e8t - 5 + C
This is almost the general solution. It would be y(t) = Ae2t + 5e8t - 5.
The first part is the complementary solution (the solution to the homogeneous diff. equation) -- what you're calling the null solution. The second part is the particular solution, a solution to the nonhomogeneous problem.
goonking said:
and then find the derivative of both sides

y' = 2e2t + 40e8t

is this correct?
No, and it's nowhere near close.
What you're supposed to do is find the nonhomogeneous differential equation that has y(t) as its solution.Your textbook should have some examples that are similar to this problem. That should be your first resource.
 
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