What is the solution for 2y^{''}+2y+y=0,y(0)=1,y^'(0)=1 ?

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

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2y^{''}+2y+y=0,y(0)=1,y^'(0)=1

2 y^{^{''}} + 2 y + y = 0, y (0) = 1, y^{^{'}} (0) = 1

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

Solution steps

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

Solve linear ODE:

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

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

0.05 q^{^{''}} + 2 q^{^{'}} + \frac{1}{0.01} q = 0

x^{^{''}} + 6 x^{^{'}} + 25 x = 0, x (0) = 3, x^{^{'}} (0) = 8

\frac{d ^{2} y}{d x ^{2}} = k y

y^{^{''}} - 12 y^{^{'}} + 20 y = 0

3 y^{^{''}} + 12 y^{^{'}} + 39 y = 0

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