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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int_{1}^{5} 5 x - \frac{5}{x} d x

\int \frac{x ^{5}}{( 2 - x ^{6} ) ^{2}} d x

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

n \to \infty lim (\frac{( cos ( \frac{1}{n} ) - 1 )}{\frac{1}{n}})

\int - 5 t d t

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