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

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SUMMARY

The discussion focuses on solving the trigonometric differential equation y'' + 36y = 0 with initial conditions y(0) = -4 and y(π/12) = 3. The auxiliary equation r² + 36 = 0 yields complex roots ±6i, leading to the general solution y = c1cos(6x) + c2sin(6x). The user initially misapplied the initial conditions, confusing y and y' evaluations, which resulted in incorrect values for c1 and c2. Correctly applying the boundary conditions to the function y will yield the appropriate constants.

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  • Understanding of second-order linear differential equations
  • Familiarity with trigonometric functions and their derivatives
  • Knowledge of initial value problems in differential equations
  • Ability to solve complex numbers and apply them in differential equations
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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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