What is the solution for y^{''}+9y=0,y(0)=2,y^'(0)=1 ?

The solution for y^{''}+9y=0,y(0)=2,y^'(0)=1 is y=2cos(3t)+1/3 sin(3t)

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y^{''}+9y=0,y(0)=2,y^'(0)=1

y^{^{''}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = 1

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

Solution steps

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

Solve linear ODE:

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

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

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

3 x^{^{''}} + 25 x = 0, x (0) = - 0.5, x^{^{'}} (0) = 2

y^{^{''}} + 6 y^{^{'}} + 8.96 y = 0

y^{^{''}} + 2 y^{^{'}} + 1.25 y = 0

y^{^{''}} + 20 y^{^{'}} + 36 y = 0, y (0) = 3, y^{^{'}} (0) = 0

What is the solution for y^{''}+9y=0,y(0)=2,y^'(0)=1 ?
The solution for y^{''}+9y=0,y(0)=2,y^'(0)=1 is y=2cos(3t)+1/3 sin(3t)