What is the solution for y^{''}+8y=0 ?

The solution for y^{''}+8y=0 is y=c_{1}cos(2^{1+1/2}t)+c_{2}sin(2^{1+1/2}t)

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y^{''}+8y=0

y^{^{''}} + 8 y = 0

y = c_{1} cos (2^{1 + \frac{1}{2}} t) + c_{2} sin (2^{1 + \frac{1}{2}} t)

Solution steps

y^{''} (t) + 8 y = 0

Solve linear ODE:

y = c_{1} cos (2^{1 + \frac{1}{2}} t) + c_{2} sin (2^{1 + \frac{1}{2}} t)

y = c_{1} cos (2^{1 + \frac{1}{2}} t) + c_{2} sin (2^{1 + \frac{1}{2}} t)

\frac{d}{d x} (arctan (9^{x}))

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

\frac{\partial}{\partial x} (3 - (x^{2} + y^{2})^{\frac{1}{3}})

\int 7 e^{- 0.4 x} d x

(2 y + 1) = \frac{d y}{d t}

What is the solution for y^{''}+8y=0 ?
The solution for y^{''}+8y=0 is y=c_{1}cos(2^{1+1/2}t)+c_{2}sin(2^{1+1/2}t)