What is the solution for y^{''}+6y^'+25y=0,y(0)=4,y^'(0)=9 ?

The solution for y^{''}+6y^'+25y=0,y(0)=4,y^'(0)=9 is y=e^{-3t}(4cos(4t)+21/4 sin(4t))

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y^{''}+6y^'+25y=0,y(0)=4,y^'(0)=9

y^{^{''}} + 6 y^{^{'}} + 25 y = 0, y (0) = 4, y^{^{'}} (0) = 9

y = e^{- 3 t} (4 cos (4 t) + \frac{21}{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{21}{4} sin (4 t))

y = e^{- 3 t} (4 cos (4 t) + \frac{21}{4} sin (4 t))

y^{^{'''}} + 11 y^{^{''}} + 38 y^{^{'}} + 40 y = 0

y^{^{''}} + 14 y^{^{'}} = - 49 y

x^{^{''}} - k^{2} x = 0

y^{^{''}} + 6 y^{^{'}} + 13 y = 0, y (0) = 2, y^{^{'}} (0) = 0

y^{^{''}} + 16 y = 0, y (\frac{π}{2}) = - 4, y^{^{'}} (\frac{π}{2}) = 8

What is the solution for y^{''}+6y^'+25y=0,y(0)=4,y^'(0)=9 ?
The solution for y^{''}+6y^'+25y=0,y(0)=4,y^'(0)=9 is y=e^{-3t}(4cos(4t)+21/4 sin(4t))