Auxiliary Eqns with Complex Roots

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Homework Help Overview

The problem involves solving an initial value problem for a second-order linear differential equation, specifically \(y'' + 9y = 0\) with initial conditions \(y(0) = 1\) and \(y'(0) = 1\). The subject area is differential equations, focusing on the characteristics of solutions with complex roots.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the characteristic equation and the roots, noting that they are complex. There is an attempt to confirm the general solution form and to evaluate the implications of the roots on the solution. Questions arise regarding the handling of double roots and the correctness of algebraic manipulations related to the derivatives.

Discussion Status

The discussion is ongoing, with some participants confirming the general solution form while others express uncertainty about the implications of double roots and the correctness of derived expressions. There is no explicit consensus, but guidance has been offered regarding the general solution.

Contextual Notes

Participants are navigating the complexities of the solution process, particularly concerning the initial conditions and the nature of the roots. There is mention of algebraic errors that may affect the interpretation of the solution.

marco101
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Homework Statement


solve the initial value problem
y''+9y = 0, y(0) = 1, y'(0) = 1


Homework Equations


gen solution form is y(t) = C1e^At*(cosBt) + C2e^At*(sinBt)
where A is the real number and B is the imaginary number

The Attempt at a Solution


i just wanted to check if I am doing this right.
i started off finding the roots,
r^2 + 9
r = +/- 3i
since the real is 0, e would be raised to 0 and become one leaving me with C1cos3t + C2sin3t
 
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marco101 said:

The Attempt at a Solution


i just wanted to check if I am doing this right.
i started off finding the roots,
r^2 + 9
r = +/- 3i
since the real is 0, e would be raised to 0 and become one leaving me with C1cos3t + C2sin3t

Yes that is correct. Now find y(0) and y'(0)
 
since there are double roots, should i make changes to the general solution form? everything cancels out to be zero because y'' = -9cos3t -9sin3t and y = 9cos3t + 9sin3t because it was multiplied by 9 in the original equation.
 
marco101 said:
since there are double roots, should i make changes to the general solution form? everything cancels out to be zero because y'' = -9cos3t -9sin3t and y = 9cos3t + 9sin3t because it was multiplied by 9 in the original equation.

double roots would mean r=3i,3i or r=-3i,-3i. You have r=0±3i. So yes your general solution is correct.

Also from y = 9cos3t + 9sin3t, y''≠ y'' = -9cos3t -9sin3t , recheck your algebra on that one.

Anyway you take it y= 9cos3t +9sin3t is correct.
 

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