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)

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)