What is the solution for y^{''}+9y=0,y(0)=5,y^'(0)=2 ?

The solution for y^{''}+9y=0,y(0)=5,y^'(0)=2 is y=5cos(3t)+2/3 sin(3t)

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y^{''}+9y=0,y(0)=5,y^'(0)=2

y^{^{''}} + 9 y = 0, y (0) = 5, y^{^{'}} (0) = 2

y = 5 cos (3 t) + \frac{2}{3} sin (3 t)

Solution steps

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

Solve linear ODE:

y = 5 cos (3 t) + \frac{2}{3} sin (3 t)

y = 5 cos (3 t) + \frac{2}{3} sin (3 t)

u^{2} y^{^{''}} - 4 u y^{^{'}} + 6 y = 0

9 y^{^{''}} + 30 y = 0, y (0) = 8, y^{^{'}} (0) = 8

50 x^{^{''}} + 100 x^{^{'}} + 1000 x = 0

\frac{1}{10} y^{^{''}} + 11 y^{^{'}} + 24 y = 0, y (0) = 1, y^{^{'}} (0) = 0

q^{^{''}} + 6 q^{^{'}} + 9 q = 0

What is the solution for y^{''}+9y=0,y(0)=5,y^'(0)=2 ?
The solution for y^{''}+9y=0,y(0)=5,y^'(0)=2 is y=5cos(3t)+2/3 sin(3t)