What is the solution for y^{''}+6y^'+25y=0,y(0)=3,y^'(0)=5 ?

The solution for y^{''}+6y^'+25y=0,y(0)=3,y^'(0)=5 is y=e^{-3t}(3cos(4t)+7/2 sin(4t))

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y^{''}+6y^'+25y=0,y(0)=3,y^'(0)=5

y^{^{''}} + 6 y^{^{'}} + 25 y = 0, y (0) = 3, y^{^{'}} (0) = 5

y = e^{- 3 t} (3 cos (4 t) + \frac{7}{2} sin (4 t))

Solution steps

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

Solve linear ODE:

y = e^{- 3 t} (3 cos (4 t) + \frac{7}{2} sin (4 t))

y = e^{- 3 t} (3 cos (4 t) + \frac{7}{2} sin (4 t))

36 y^{^{''}} + 108 y^{^{'}} + 97 y = 0

y^{^{''}} + 4 y^{^{'}} - 3 y = 0

y^{^{''}} + 2 y^{^{'}} + 6 y = 0, y (0) = 1, y^{^{'}} (0) = - 2

y^{^{''}} - 3 y^{^{'}} - 4 y = 0, y (0) = a, y^{^{'}} (0) = - 15

2 y^{^{''}} + 128 y = 0

What is the solution for y^{''}+6y^'+25y=0,y(0)=3,y^'(0)=5 ?
The solution for y^{''}+6y^'+25y=0,y(0)=3,y^'(0)=5 is y=e^{-3t}(3cos(4t)+7/2 sin(4t))