# Differential Equation

## Homework Statement

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

## 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 cancelled out?).

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

Thanks for any help

Mark44
Mentor

## Homework Statement

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

## 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 cancelled 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

phyzguy