Brushing up on the basics of diff eq

[cowmoo32]

Homework Statement

y(x) = C₁sin(2x) + C₂cos(2x)

Homework Equations

y(∏/8) = 0

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

 

[SammyS]

Homework Statement

y(x) = C₁sin(2x) + C₂cos(2x)

Homework Equations

y(∏/8) = 0

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

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

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

 

[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

 

[SammyS]

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

 

[cowmoo32]

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

 

[SammyS]

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

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 $-C_1-\sqrt{2}$