What is the solution for 3y^{''}+27y=0,y(0)=0,y^'(0)=1 ?

The solution for 3y^{''}+27y=0,y(0)=0,y^'(0)=1 is y= 1/3 sin(3t)

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3y^{''}+27y=0,y(0)=0,y^'(0)=1

3 y^{^{''}} + 27 y = 0, y (0) = 0, y^{^{'}} (0) = 1

y = \frac{1}{3} sin (3 t)

Solution steps

3 y^{''} (t) + 27 y = 0

Solve linear ODE:

y = \frac{1}{3} sin (3 t)

y = \frac{1}{3} sin (3 t)

y^{^{''}} + 4 y = 0, y (0) = 1, y^{^{'}} (0) = 1

9 y^{^{''}} - 6 y^{^{'}} = 0

y^{^{'''}} + 4 y^{^{''}} = 0

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = 0, y^{^{'}} (0) = 2

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

What is the solution for 3y^{''}+27y=0,y(0)=0,y^'(0)=1 ?
The solution for 3y^{''}+27y=0,y(0)=0,y^'(0)=1 is y= 1/3 sin(3t)