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

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

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

linear

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

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

Solution steps

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

Solve linear ODE:

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

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

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

(y^{^{''}} - a y^{^{'}}) = b y

y^{^{''}} + 12 y^{^{'}} + 4 y = 0, y (0) = - 3, y^{^{'}} (0) = 0

y^{^{''}} + 16.2 y^{^{'}} = 0

(D^{2} + 4 D + 4) y = 0

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