How Do You Solve a Trigonometric Differential Equation with Initial Conditions?

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Homework Help Overview

The discussion revolves around solving a trigonometric differential equation of the form y'' + 36y = 0, with specific initial conditions y(0) = -4 and y(π/12) = 3. Participants are exploring how to find the particular solution that satisfies these conditions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss setting up the auxiliary equation and finding the general solution involving constants c1 and c2. There are questions about the differentiation process and the application of initial conditions to determine these constants. Some participants express confusion over the evaluation of boundary values and the roles of y and y' in this context.

Discussion Status

The discussion is active, with participants providing insights and corrections to each other's approaches. There is recognition of a mistake regarding the application of boundary conditions, and some guidance has been offered on how to correctly evaluate the constants. However, there is still uncertainty regarding the values of c1 and c2.

Contextual Notes

Participants are working under the constraints of the initial conditions provided and are questioning the setup of their equations. There is a noted mix-up between the variables used in the solution process.

DanielJackins
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Homework Statement



Find the particular solution of the differential equation y''+36y=0 satisfying the conditions y(0)=−4 and y(pi/12)=3.

Your answer should be a function of x.

The Attempt at a Solution



I think I know how to do this kind of question, and I can't see where I'm going wrong. I set up the auxiliary equation, r^2+36 = 0, and get the roots to be + or - 6i. So then I get y = c1cos(6x)+c2sin(6t). I differentiate, and get y' = -6c1sin(6x)+6c2cos(6x). Then I plug in the initial values given. The first equation gives me c1 = -4, and the second gives me 3 = -6c1, which is clearly incorrect (and in both equations c2 is canceled out?).

I've looked over it over and over again but I can't see where I'm going wrong.

Thanks for any help
 
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DanielJackins said:

Homework Statement



Find the particular solution of the differential equation y''+36y=0 satisfying the conditions y(0)=−4 and y(pi/12)=3.

Your answer should be a function of x.

The Attempt at a Solution



I think I know how to do this kind of question, and I can't see where I'm going wrong. I set up the auxiliary equation, r^2+36 = 0, and get the roots to be + or - 6i. So then I get y = c1cos(6x)+c2sin(6t). I differentiate, and get y' = -6c1sin(6x)+6c2cos(6x). Then I plug in the initial values given. The first equation gives me c1 = -4, and the second gives me 3 = -6c1, which is clearly incorrect (and in both equations c2 is canceled out?).

I've looked over it over and over again but I can't see where I'm going wrong.

Thanks for any help
Your solution is correct (but don't mix x and t). Just evaluate your solution at the two boundary points to find c1 and c2.
 
But when I evaluate to find c1 and c2 I seem to get two different values for c1, and no value for c2? I'm not sure I understand
 
Why are you putting the second boundary value into the equation for y'? Both boundary values are for y, not y'. One will fix C1 and the other will fix C2.
 
Oh man, didn't even notice that! Thanks!
 

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