What is the solution for y^{''}+6y^'+25y=0,y(0)=4,y^'(0)=-11 ?

The solution for y^{''}+6y^'+25y=0,y(0)=4,y^'(0)=-11 is y=e^{-3t}(4cos(4t)+1/4 sin(4t))

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y^{''}+6y^'+25y=0,y(0)=4,y^'(0)=-11

y^{^{''}} + 6 y^{^{'}} + 25 y = 0, y (0) = 4, y^{^{'}} (0) = - 11

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

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

64 y^{^{''}} - 81 y = 0

y^{^{''}} + 9 y = 0, y (0) = 3, y^{^{'}} (0) = - 4

y^{^{''}} - 6 y^{^{'}} + 25 y = 0, y (0) = 2, y^{^{'}} (0) = 2

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

y^{^{''}} - 2 y^{^{'}} + 2 y = 0, y^{^{'}} (π) = 0, y (1) = 0

What is the solution for y^{''}+6y^'+25y=0,y(0)=4,y^'(0)=-11 ?
The solution for y^{''}+6y^'+25y=0,y(0)=4,y^'(0)=-11 is y=e^{-3t}(4cos(4t)+1/4 sin(4t))