What is the solution for y^{''}+y=0,y(pi/3)=0,y^'(pi/3)=8 ?

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

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y^{''}+y=0,y(pi/3)=0,y^'(pi/3)=8

y^{^{''}} + y = 0, y (\frac{π}{3}) = 0, y^{^{'}} (\frac{π}{3}) = 8

y = - 3 \cdot 4 cos (t) + 4 sin (t)

Solution steps

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

Solve linear ODE:

y = - 3 \cdot 4 cos (t) + 4 sin (t)

y = - 3 \cdot 4 cos (t) + 4 sin (t)

\frac{\partial}{\partial x} (3 x + 9 x^{17})

x \to 0 - lim (x^{3} cos (\frac{2}{x}))

\frac{d}{d x} (\frac{e ^{x}}{4 x ^{3}})

\int \frac{1}{1 - tan ( x )} d x

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