# Differential Equation

• DanielJackins

## 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 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

## 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 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!