What is the solution for y^{''}+2y+y=0 ?

The solution for y^{''}+2y+y=0 is y=c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t)

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y^{''}+2y+y=0

y^{^{''}} + 2 y + y = 0

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

Solution steps

y^{''} (t) + 2 y + y = 0

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)

f (x) = (\frac{5}{x ^{2} + 1}), at x = - 1

y^{^{'}} = 2 e^{(6 x - y)}

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

\frac{\partial}{\partial z} (2 yz)

f (x) = x - 4 tan (x), at x = \frac{3 π}{4}

What is the solution for y^{''}+2y+y=0 ?
The solution for y^{''}+2y+y=0 is y=c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t)