What is the solution for y^{''}+3y=0,y^0=1,y^'0=3 ?

The solution for y^{''}+3y=0,y^0=1,y^'0=3 is y=c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t)

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y^{''}+3y=0,y^0=1,y^'0=3

y^{^{''}} + 3 y = 0, y^{0} (0) = 1, y^{^{'}} 0 = 3

y = c_{1} cos (3 t) + c_{2} sin (3 t)

Solution steps

y^{''} (t) + 3 y = 0

Solve linear ODE:

y = c_{1} cos (3 t) + c_{2} sin (3 t)

y = c_{1} cos (3 t) + c_{2} sin (3 t)

\frac{1}{2} y^{^{''}} + 6 y^{^{'}} + 26 y = 0

y^{^{''}} - 9 y^{^{'}} = 0, y (0) = e^{3 x}

y^{^{''}} - 2 y^{^{'}} + 50 y = 0

y^{^{'''}} + 4 y^{^{''}} - 15 y^{^{'}} - 18 y = 0

y^{^{''}} + 2 y^{^{'}} + 10 y = 0, y (0) = 3, y^{^{'}} (0) = 6

What is the solution for y^{''}+3y=0,y^0=1,y^'0=3 ?
The solution for y^{''}+3y=0,y^0=1,y^'0=3 is y=c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t)