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

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

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y^{''}+6y^'+25y=0,y(0)=a,y^'(0)=b

y^{^{''}} + 6 y^{^{'}} + 25 y = 0, y (0) = a, y^{^{'}} (0) = b

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

Solution steps

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

Solve linear ODE:

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

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

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