What is the solution for 16y^{''}+40y^'+169y=0,y(0)=8,y^'(0)=18 ?

The solution for 16y^{''}+40y^'+169y=0,y(0)=8,y^'(0)=18 is y=e^{-(5t)/4}(8cos(3t)+28/3 sin(3t))

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16y^{''}+40y^'+169y=0,y(0)=8,y^'(0)=18

16 y^{^{''}} + 40 y^{^{'}} + 169 y = 0, y (0) = 8, y^{^{'}} (0) = 18

y = e^{- \frac{5 t}{4}} (8 cos (3 t) + \frac{28}{3} sin (3 t))

Solution steps

16 y^{''} (t) + 40 y^{'} (t) + 169 y = 0

Solve linear ODE:

y = e^{- \frac{5 t}{4}} (8 cos (3 t) + \frac{28}{3} sin (3 t))

y = e^{- \frac{5 t}{4}} (8 cos (3 t) + \frac{28}{3} sin (3 t))

What is the solution for 16y^{''}+40y^'+169y=0,y(0)=8,y^'(0)=18 ?
The solution for 16y^{''}+40y^'+169y=0,y(0)=8,y^'(0)=18 is y=e^{-(5t)/4}(8cos(3t)+28/3 sin(3t))