What is the solution for y^{''}+6y^'+25y=0,y(0)=4,y^'(0)=5 ?

The solution for y^{''}+6y^'+25y=0,y(0)=4,y^'(0)=5 is y=e^{-3t}(4cos(4t)+17/4 sin(4t))

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y^{''}+6y^'+25y=0,y(0)=4,y^'(0)=5

y^{^{''}} + 6 y^{^{'}} + 25 y = 0, y (0) = 4, y^{^{'}} (0) = 5

y = e^{- 3 t} (4 cos (4 t) + \frac{17}{4} sin (4 t))

Solution steps

y^{''} (t) + 6 y^{'} (t) + 25 y = 0

Solve linear ODE:

y = e^{- 3 t} (4 cos (4 t) + \frac{17}{4} sin (4 t))

y = e^{- 3 t} (4 cos (4 t) + \frac{17}{4} sin (4 t))

2 y^{^{''}} + 50 y = 0, y (0) = 8, y^{^{'}} (0) = 35

y^{^{''}} + 9 y^{^{'}} = 0, y (0) = 3, y^{^{'}} (0) = - 45

y^{^{''}} + \frac{33}{6} y^{^{'}} + 128 y = 0

15 y^{^{''}} - 16 y^{^{'}} - 7 y = 0

y^{^{''}} - 11 y^{^{'}} - 42 y = 0

What is the solution for y^{''}+6y^'+25y=0,y(0)=4,y^'(0)=5 ?
The solution for y^{''}+6y^'+25y=0,y(0)=4,y^'(0)=5 is y=e^{-3t}(4cos(4t)+17/4 sin(4t))