Finding a linear differential equation

In summary, the conversation discusses finding the general solution to a nonhomogeneous differential equation. The general solution is the sum of the complementary solution and the particular solution. The complementary solution is also referred to as the null solution. The particular solution is a solution to the nonhomogeneous problem. The correct general solution is y(t) = Ae2t + 5e8t - 5. The next step is to find the nonhomogeneous differential equation that has y(t) as its solution. It is recommended to refer to the textbook for examples.
  • #1
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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  • #2
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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