What is the solution for y^{''}+6y^'+25y=0,y(0)=1,y^'(0)=-1 ?

The solution for y^{''}+6y^'+25y=0,y(0)=1,y^'(0)=-1 is y=e^{-3t}(cos(4t)+1/2 sin(4t))

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y^{''}+6y^'+25y=0,y(0)=1,y^'(0)=-1

y^{^{''}} + 6 y^{^{'}} + 25 y = 0, y (0) = 1, y^{^{'}} (0) = - 1

y = e^{- 3 t} (cos (4 t) + \frac{1}{2} sin (4 t))

Solution steps

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

Solve linear ODE:

y = e^{- 3 t} (cos (4 t) + \frac{1}{2} sin (4 t))

y = e^{- 3 t} (cos (4 t) + \frac{1}{2} sin (4 t))

\frac{x ^{^{''}} + ( n - 1 ) x ^{^{'}}}{r} = 0

(D^{3} - 4 D) y = 0

y^{^{''}} + 10 y^{^{'}} - 11 y = 0

(4 D^{3} - 13 D - 6) y = 0

r^{2} x^{^{''}} + r x^{^{'}} - n^{2} x = 0

What is the solution for y^{''}+6y^'+25y=0,y(0)=1,y^'(0)=-1 ?
The solution for y^{''}+6y^'+25y=0,y(0)=1,y^'(0)=-1 is y=e^{-3t}(cos(4t)+1/2 sin(4t))