Therefore, y(t) = k1y1(t) + k2y2(t) is a solution to the given DE.

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Discussion Overview

The discussion revolves around verifying that a linear combination of two solutions to a second-order linear differential equation (DE) is also a solution. Participants explore the theoretical aspects of this verification, particularly in the context of a homework problem involving arbitrary functions.

Discussion Character

  • Homework-related
  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion about verifying the solution for arbitrary functions rather than specific examples.
  • Another participant suggests using the linearity principle and plugging the linear combination into the original DE to check for equality.
  • A later reply indicates that solving for the second derivative in the context of the original equations leads to a cancellation that supports the verification.
  • Further clarification is provided on how inserting the linear combination into the DE results in an equality of zero, reinforcing the solution's validity.

Areas of Agreement / Disagreement

Participants generally agree on the approach to verify the solution using the linearity principle, but the discussion remains focused on the process rather than reaching a definitive conclusion about the verification method itself.

Contextual Notes

Participants discuss the verification process without resolving all assumptions about the functions involved or the specific conditions under which the linearity principle applies.

KevinL
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I understand how to solve a normal second order linear equation, but this question in the homework is a bit more theoretical and I'm a bit confused.

"Suppose y1(t) and y2(t) are solutions of y'' + py' + qy = 0

Verify that y(t) = k1y1(t) + k2y2(t) is also a solution for any choice of constants k1 and k2."

If we were given actual functions (i.e. e^-t) this would be very simple. Just plug it into the DE and see that you have an equality. But when given arbitrary functions I don't know how to verify that its an equality. I am guessing it has something to do with the Linearity Principle, but I don't understand how to actually verify it.
 
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Since it's homework, I probably shouldn't give the answer. But a hint...

Since y_1 and y_2 are solutions, then

y1'' + py1' + qy1 = 0
y2'' + py2' + qy2 = 0

How about plugging (k1y1 + k2y2) into the original equation and rearranging a bit?
 
That got me started on the right track, I think. With the two equations you have, solve each one for y''. Then plug that into the other equation and everything nicely cancels so that 0=0.
 
Sounds like you have it. You just need to show that inserting y into the equation will equal zero as well.

(k1y1 + k2y2)'' + p(k1y1 + k2y2)' + q(k1y1 + k2y2) = 0

k1y1'' + k2y2'' + pk1y1' + pk2y2' + qk1y1 + qk2y2 = 0

k1(y1'' + py1' + qy1) + k2(y2'' + py2' + qy2) = 0

k1*0 + k2*0 = 0 ==> 0 = 0
 

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