What is the solution for y^{''}+16y^'+73y=0,y^'(0)=8 ?

The solution for y^{''}+16y^'+73y=0,y^'(0)=8 is y=e^{-8t}(-(cos(3t)(8-3c_{2}))/8+c_{2}sin(3t))

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y^{''}+16y^'+73y=0,y^'(0)=8

y^{^{''}} + 16 y^{^{'}} + 73 y = 0, y^{^{'}} (0) = 8

y = e^{- 8 t} (- \frac{cos ( 3 t ) ( 8 - 3 c _{2} )}{8} + c_{2} sin (3 t))

Solution steps

y^{''} (t) + 16 y^{'} (t) + 73 y = 0

Solve linear ODE:

y = e^{- 8 t} (- \frac{cos ( 3 t ) ( 8 - 3 c _{2} )}{8} + c_{2} sin (3 t))

y = e^{- 8 t} (- \frac{cos ( 3 t ) ( 8 - 3 c _{2} )}{8} + c_{2} sin (3 t))

y^{^{''}} + 8 y^{^{'}} + 20 y = 0, y (0) = 1, y^{^{'}} (0) = 4

y^{^{''}} + 8 y^{^{'}} + 20 y = 0, y (0) = 1, y^{^{'}} (0) = 7

x^{^{''}} - x^{^{'}} - 110 x = 0, x (0) = 14

y^{^{''}} - 6 y^{^{'}} - 40 y = 0

2 y^{^{''}} - y^{^{'}} - 3 y = 0, y (0) = 0, y^{^{'}} (0) = 5

What is the solution for y^{''}+16y^'+73y=0,y^'(0)=8 ?
The solution for y^{''}+16y^'+73y=0,y^'(0)=8 is y=e^{-8t}(-(cos(3t)(8-3c_{2}))/8+c_{2}sin(3t))