What is the solution for y^{''}+4y=0,y^'(0)=2 ?

The solution for y^{''}+4y=0,y^'(0)=2 is y=c_{1}cos(2t)+sin(2t)

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y^{''}+4y=0,y^'(0)=2

y^{^{''}} + 4 y = 0, y^{^{'}} (0) = 2

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

Solution steps

y^{''} (t) + 4 y = 0

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} + \frac{1}{4} y = 0

100 y^{^{''}} + 200 y^{^{'}} + 101 y = 0

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

\frac{d ^{2} x}{d t ^{2}} + 8 \frac{d x}{d t} + 4 x = 0

9 y^{^{''}} - 4 y = 0

What is the solution for y^{''}+4y=0,y^'(0)=2 ?
The solution for y^{''}+4y=0,y^'(0)=2 is y=c_{1}cos(2t)+sin(2t)