What is the solution for y^{''}+18y^'+117y=0,y(0)=6,y^'(0)=3 ?

The solution for y^{''}+18y^'+117y=0,y(0)=6,y^'(0)=3 is y=e^{-9t}(6cos(6t)+19/2 sin(6t))

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y^{''}+18y^'+117y=0,y(0)=6,y^'(0)=3

y^{^{''}} + 18 y^{^{'}} + 117 y = 0, y (0) = 6, y^{^{'}} (0) = 3

y = e^{- 9 t} (6 cos (6 t) + \frac{19}{2} sin (6 t))

Solution steps

y^{''} (t) + 18 y^{'} (t) + 117 y = 0

Solve linear ODE:

y = e^{- 9 t} (6 cos (6 t) + \frac{19}{2} sin (6 t))

y = e^{- 9 t} (6 cos (6 t) + \frac{19}{2} sin (6 t))

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

20 y^{^{''}} + 27 y^{^{'}} + 9 y = 0, y (0) = 0, y^{^{'}} (0) = 4

4 y^{^{''}} + y^{^{'}} - 3 y = 0, y (0) = - 7

y^{^{''}} + 49 π^{2} y = 0

\frac{d ^{2} x}{d t ^{2}} + 15 \frac{d x}{d t} + 56 x = 0

What is the solution for y^{''}+18y^'+117y=0,y(0)=6,y^'(0)=3 ?
The solution for y^{''}+18y^'+117y=0,y(0)=6,y^'(0)=3 is y=e^{-9t}(6cos(6t)+19/2 sin(6t))