# Find a solution to the boundary value problem y'' +4y = 0 ; y(π/8)=0 , y(π/6)=1, if the gen. soln. to the D.E is y(x)=C1sin(2x) + C2cos(2x) ?

I'm currently in case 1, where y(π/8) = 0.

I substituted y(π/8) = 0 to y(x)=C1sin(2x) + C2cos(2x) and got the value for constant which C1 + C2 = 0. But I can't find a way to substitute this back into the equation to get a particular solution.

Pls. help. Or hint if there is something wrong with my method.

thanks, I'm currently new to this so I don't know what to do in the Boundary value problem where there are two conditions. I'll try again.

I got it now thanks.

### 2 Answers

- 2 months ago
y(pi/8) = 0

y(pi/6) = 1

0 = C1 * sin(2 * pi/8) + C2 * cos(2 * pi/8)

1 = C1 * sin(2 * pi/6) + C2 * cos(2 * pi/6)

0 = C1 * sin(pi/4) + C2 * cos(pi/4)

1 = C1 * sin(pi/3) + C2 * cos(pi/3)

0 = C1 * (sqrt(2)/2) + C2 * (sqrt(2)/2)

1 = C1 * (sqrt(3)/2) + C2 * (1/2)

0 = C1 + C2

1 = C1 * sqrt(3) + C2

1 - 0 = C1 * sqrt(3) - C1 + C2 - C2

1 = C1 * (sqrt(3) - 1) + 0 * C2

1 = C1 * (sqrt(3) - 1)

1 / (sqrt(3) - 1) = C1

(sqrt(3) + 1) / (3 - 1) = C1

(sqrt(3) + 1) / 2 = C1

0 = C1 + C2

C2 = -C1

C2 = -(sqrt(3) + 1) / 2