# Brushing up on the basics of diff eq

## Homework Statement

y(x) = C1sin(2x) + C2cos(2x)

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
Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

y(x) = C1sin(2x) + C2cos(2x)

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 .

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

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

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