What is the solution for 16y^{''}+8y^'+145y=0,y(0)=-2,y^'(0)=1 ?

The solution for 16y^{''}+8y^'+145y=0,y(0)=-2,y^'(0)=1 is y=e^{-t/4}(-2cos(3t)+1/6 sin(3t))

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16y^{''}+8y^'+145y=0,y(0)=-2,y^'(0)=1

16 y^{^{''}} + 8 y^{^{'}} + 145 y = 0, y (0) = - 2, y^{^{'}} (0) = 1

y = e^{- \frac{t}{4}} (- 2 cos (3 t) + \frac{1}{6} sin (3 t))

Solution steps

16 y^{''} (t) + 8 y^{'} (t) + 145 y = 0

Solve linear ODE:

y = e^{- \frac{t}{4}} (- 2 cos (3 t) + \frac{1}{6} sin (3 t))

y = e^{- \frac{t}{4}} (- 2 cos (3 t) + \frac{1}{6} sin (3 t))

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

9 (\frac{d ^{2} y}{d t ^{2}}) - 24 (\frac{d y}{d t}) + 16 y = 0

x^{^{''}} + 64 x = 0, x (0) = \frac{4}{12}, x^{^{'}} (0) = 0

y^{^{'''}} - 2 y^{^{''}} + 8 y^{^{'}} = 0

y^{^{''}} + 16 y = 0, y^{^{'}} (0) = 3, y (0) = 5

What is the solution for 16y^{''}+8y^'+145y=0,y(0)=-2,y^'(0)=1 ?
The solution for 16y^{''}+8y^'+145y=0,y(0)=-2,y^'(0)=1 is y=e^{-t/4}(-2cos(3t)+1/6 sin(3t))