# Brushing up on the basics of diff eq

1. Dec 30, 2011

### cowmoo32

1. The problem statement, all variables and given/known data
y(x) = C1sin(2x) + C2cos(2x)

2. Relevant equations
y(∏/8) = 0

3. The attempt at a solution
C1(1/2)(√2) + C2(1/2)(√2) + 1 = 0

The book jumps to the equation below and I'm having trouble figuring out how they got there.
C1 + C2 = -√2

2. Dec 30, 2011

### SammyS

Staff Emeritus
What is the dif. eq. you're trying to solve?

Multiply your "attempt at a solution" by √2 .

3. Dec 30, 2011

### cowmoo32

Determine C1 and C2 so that y(x) = C1sin(2x) + C2cos(2x) will satisfy the initial conditions y(∏/8) = 0 and y'(∏/8) = √2

4. Dec 30, 2011

### SammyS

Staff Emeritus
Multiply your "attempt at a solution" in your original post by √2 .

5. Dec 30, 2011

### cowmoo32

I guess it's safe to assume that C1/2 = C1 and C2/2 = C2 considering they're both unknowns at this point?

6. Dec 30, 2011

### SammyS

Staff Emeritus
That doesn't make sense, unless you are asking if you can replace C1/2 by C1 and C2/2 by C2 .

If $C_1+C_2=-\sqrt{2}\,,$ then you can replace $C_2$ with $-C_1-\sqrt{2}$