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

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

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

linear

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

y = cos (t)

Solution steps

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

Solve linear ODE:

y = cos (t)

y = cos (t)

\frac{y ^{^{''}}}{8} + 30 y^{^{'}} + 500 y = 0

y^{^{''}} - 36 \cdot y = 0

y^{^{''}} - 23 y^{^{'}} + 112 y = 0, y (0) = 3, y^{^{'}} (0) = 48

y^{^{'''}} + 8 y^{^{''}} = 0

f^{^{''}} (t) + f (t) = 0, f^{^{'}} (0) = 1

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