What is the solution for 4y^{''}+8y=0 ?

The solution for 4y^{''}+8y=0 is y=c_{1}cos(sqrt(2)t)+c_{2}sin(sqrt(2)t)

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4y^{''}+8y=0

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

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

Solution steps

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

Solve linear ODE:

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

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

x \to - 1 lim (sin (π x))

\frac{d}{d x} (2 x y^{3})

a re a x^{3} - 10 x^{2} + 16 x, - x^{3} + 10 x^{2} - 16 x

\frac{d}{d x} ((x^{2} - 5 x + 3) (4 x^{3} + 10 x - 5))

e x t re m e f (x) = x^{2} - 11 y^{2}

What is the solution for 4y^{''}+8y=0 ?
The solution for 4y^{''}+8y=0 is y=c_{1}cos(sqrt(2)t)+c_{2}sin(sqrt(2)t)