What is the solution for y^{''}+6y^'+25y=0,y(0)=3,y^'(0)=-8 ?

The solution for y^{''}+6y^'+25y=0,y(0)=3,y^'(0)=-8 is y=e^{-3t}(3cos(4t)+1/4 sin(4t))

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y^{''}+6y^'+25y=0,y(0)=3,y^'(0)=-8

y^{^{''}} + 6 y^{^{'}} + 25 y = 0, y (0) = 3, y^{^{'}} (0) = - 8

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

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

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What is the solution for y^{''}+6y^'+25y=0,y(0)=3,y^'(0)=-8 ?
The solution for y^{''}+6y^'+25y=0,y(0)=3,y^'(0)=-8 is y=e^{-3t}(3cos(4t)+1/4 sin(4t))