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)

\frac{d}{d x} (x^{x + 1})

(- 20, - 4), (- 12, - 10)

n = 1 \sum \infty \frac{2}{25} \cdot (\frac{1}{5})^{n}

6 t e^{5 z}

4^{x}

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)