What is the solution for y^{''}+y^'+9.25y=0,y^'(0)=8 ?

The solution for y^{''}+y^'+9.25y=0,y^'(0)=8 is y=e^{-t/2}((-16+6c_{2})cos(3t)+c_{2}sin(3t))

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y^{''}+y^'+9.25y=0,y^'(0)=8

y^{^{''}} + y^{^{'}} + 9.25 y = 0, y^{^{'}} (0) = 8

y = e^{- \frac{t}{2}} ((- 16 + 6 c_{2}) cos (3 t) + c_{2} sin (3 t))

Solution steps

y^{''} (t) + y^{'} (t) + 9.25 y = 0

Solve linear ODE:

y = e^{- \frac{t}{2}} ((- 16 + 6 c_{2}) cos (3 t) + c_{2} sin (3 t))

y = e^{- \frac{t}{2}} ((- 16 + 6 c_{2}) cos (3 t) + c_{2} sin (3 t))

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

y^{^{''''}} - 4 y^{^{'''}} + 20 y^{^{''}} - 64 y^{^{'}} + 64 y = 0

\frac{d ^{2} x}{d t ^{2}} + 4 \frac{d x}{d t} + 5 x = 0, x (0) = 1

9 y^{^{''}} + 8 y^{^{'}} + 7 y = 0

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

What is the solution for y^{''}+y^'+9.25y=0,y^'(0)=8 ?
The solution for y^{''}+y^'+9.25y=0,y^'(0)=8 is y=e^{-t/2}((-16+6c_{2})cos(3t)+c_{2}sin(3t))