What is the solution for y^{''}+16y^'+100y=0,y(0)=6,y^'(0)=3 ?

The solution for y^{''}+16y^'+100y=0,y(0)=6,y^'(0)=3 is y=e^{-8t}(6cos(6t)+17/2 sin(6t))

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y^{''}+16y^'+100y=0,y(0)=6,y^'(0)=3

y^{^{''}} + 16 y^{^{'}} + 100 y = 0, y (0) = 6, y^{^{'}} (0) = 3

y = e^{- 8 t} (6 cos (6 t) + \frac{17}{2} sin (6 t))

Solution steps

y^{''} (t) + 16 y^{'} (t) + 100 y = 0

Solve linear ODE:

y = e^{- 8 t} (6 cos (6 t) + \frac{17}{2} sin (6 t))

y = e^{- 8 t} (6 cos (6 t) + \frac{17}{2} sin (6 t))

θ^{^{''}} (t) = 3 θ (t)

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

4 y^{^{''}} + \frac{1}{4} y^{^{'}} + y = 0

y^{^{''}} + 2 y^{^{'}} - 15 y = 0, y (0) = x, y^{^{'}} (0) = 20

3 y^{^{''}} + 2 y^{^{'}} + 1 y = 0

What is the solution for y^{''}+16y^'+100y=0,y(0)=6,y^'(0)=3 ?
The solution for y^{''}+16y^'+100y=0,y(0)=6,y^'(0)=3 is y=e^{-8t}(6cos(6t)+17/2 sin(6t))