What is the solution for y^{''}=-3y ?

The solution for y^{''}=-3y is y=c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t)

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y^{''}=-3y

y^{^{''}} = - 3 y

y = c_{1} cos (3 t) + c_{2} sin (3 t)

Solution steps

y^{''} (t) = - 3 y

Solve linear ODE:

y = c_{1} cos (3 t) + c_{2} sin (3 t)

y = c_{1} cos (3 t) + c_{2} sin (3 t)

\frac{\partial}{\partial x} (sec (x y) + tan (x y))

x \to 4 + lim (\frac{2}{x - 4})

\int 3 x^{- 5} d x

\int (e^{x} + 2) d x

f (x) = \frac{( 8 x + x ^{2} )}{( 4 + x ) ^{2}}

What is the solution for y^{''}=-3y ?
The solution for y^{''}=-3y is y=c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t)