What is the solution for y^{''}+y=0,y(pi/3)=0,y^'(pi/3)=6 ?

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

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y^{''}+y=0,y(pi/3)=0,y^'(pi/3)=6

y^{^{''}} + y = 0, y (\frac{π}{3}) = 0, y^{^{'}} (\frac{π}{3}) = 6

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

Solution steps

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

Solve linear ODE:

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

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

\int \frac{x + 1}{x ^{2}} d x

\int \frac{1}{x ^{2} + 5 x + 6} d x

y = cos (e^{- t^{2}})

\frac{d}{d x} ((x^{2} + 2) (x^{3} + 1))

\frac{d}{d t} (0.0225 t e^{- 0.0467 t})

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