What is the solution for y^{''}+4y=0,y(pi/8)=0,y(pi/6)=1 ?

The solution for y^{''}+4y=0,y(pi/8)=0,y(pi/6)=1 is y=-(1+sqrt(3))cos(2t)+(1+sqrt(3))sin(2t)

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y^{''}+4y=0,y(pi/8)=0,y(pi/6)=1

y^{^{''}} + 4 y = 0, y (\frac{π}{8}) = 0, y (\frac{π}{6}) = 1

y = - (1 + 3) cos (2 t) + (1 + 3) sin (2 t)

Solution steps

y^{''} (t) + 4 y = 0

Solve linear ODE:

y = - (1 + 3) cos (2 t) + (1 + 3) sin (2 t)

y = - (1 + 3) cos (2 t) + (1 + 3) sin (2 t)

- 4 y^{^{''}} - 4 y^{^{'}} - 4 y = 0

y^{^{''}} + 4 y = 0, y (0) = - 1, y^{^{'}} (0) = 0

y^{^{'''}} + 6 y^{^{''}} + 11 y^{^{'}} + 6 y = 0

\frac{d ^{2} y}{d x ^{2}} + 3 \frac{d y}{d x} + y = 0

y^{^{''}} - 6 y^{^{'}} + 34 y = 0, y (0) = 5, y^{^{'}} (0) = - 5

What is the solution for y^{''}+4y=0,y(pi/8)=0,y(pi/6)=1 ?
The solution for y^{''}+4y=0,y(pi/8)=0,y(pi/6)=1 is y=-(1+sqrt(3))cos(2t)+(1+sqrt(3))sin(2t)