What is the solution for y^{''}=-2y ?

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

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y^{''}=-2y

y^{^{''}} = - 2 y

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

Solution steps

y^{''} (t) = - 2 y

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)

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

\int \frac{x ^{2} - 2 x - 1}{2 x ^{3} + 3 x ^{2} - 2 x} d x

\int \frac{x ^{2}}{e ^{5 x}} d x

\frac{8 e ^{(- 3 s)}}{s ^{3}}

\frac{1}{1 - y}

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