What is the solution for (d^2y)/(dt^2)+2y=0,y^'(0)=2 ?

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

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(d^2y)/(dt^2)+2y=0,y^'(0)=2

\frac{d ^{2} y}{d t ^{2}} + 2 y = 0, y^{^{'}} (0) = 2

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = - 2, y^{^{'}} (0) = 1

\frac{1}{2} y^{^{''}} + 2 y = 0

y^{^{'''}} + 6 y^{^{''}} = 0

y^{^{'''}} - 2 y^{^{''}} + 5 y^{^{'}} = 0, y (0) = 0, y^{^{'}} (0) = 0

\frac{d ^{2} y}{d t ^{2}} - 15 \frac{d y}{d t} + 50 y = 0

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