What is the solution for y^{''}+y^'+9.25y=0,y(0)=2,y^'(0)=8 ?

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

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y^{''}+y^'+9.25y=0,y(0)=2,y^'(0)=8

y^{^{''}} + y^{^{'}} + 9.25 y = 0, y (0) = 2, y^{^{'}} (0) = 8

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{'''}} + 6 y^{^{''}} + 10 y^{^{'}} + 8 y = 0

y^{^{''}} + 2 y^{^{'}} + 26 y = 0, y (0) = 1, y^{^{'}} (0) = 4

4 y^{^{''}} - 9 y = 0, y (0) = 0, y^{^{'}} (0) = 3

y^{^{'''}} - 2 y^{^{''}} - 21 y^{^{'}} - 18 y = 0

4 x^{^{''}} + 8 x^{^{'}} + 4 x = 0

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