What is the solution for y^{''}+9y=0,y(0)=1,y^'(0)=1 ?

The solution for y^{''}+9y=0,y(0)=1,y^'(0)=1 is y=cos(3t)+1/3 sin(3t)

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y^{''}+9y=0,y(0)=1,y^'(0)=1

y^{^{''}} + 9 y = 0, y (0) = 1, y^{^{'}} (0) = 1

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

Solution steps

y^{''} (t) + 9 y = 0

Solve linear ODE:

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

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

y^{^{''}} - 14 y^{^{'}} + 45 y = 0

y^{^{''}} + 6 y^{^{'}} + 25 y = 0, y (0) = 3, y^{^{'}} (0) = 5

36 y^{^{''}} + 108 y^{^{'}} + 97 y = 0

y^{^{''}} + 4 y^{^{'}} - 3 y = 0

y^{^{''}} + 2 y^{^{'}} + 6 y = 0, y (0) = 1, y^{^{'}} (0) = - 2

What is the solution for y^{''}+9y=0,y(0)=1,y^'(0)=1 ?
The solution for y^{''}+9y=0,y(0)=1,y^'(0)=1 is y=cos(3t)+1/3 sin(3t)