What is the solution for y^{''}+8y^'+25y=0,y(0)=3,y^'(0)=-11 ?

The solution for y^{''}+8y^'+25y=0,y(0)=3,y^'(0)=-11 is y=e^{-4t}(3cos(3t)+1/3 sin(3t))

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y^{''}+8y^'+25y=0,y(0)=3,y^'(0)=-11

y^{^{''}} + 8 y^{^{'}} + 25 y = 0, y (0) = 3, y^{^{'}} (0) = - 11

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

Solution steps

y^{''} (t) + 8 y^{'} (t) + 25 y = 0

Solve linear ODE:

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

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

x^{6} + y^{5} x^{7} + 5 y^{^{'}} = 0

4 e^{- x} + e^{3 x}

\int e^{- t x} sin (x) d x

x \to - 1 lim (3 x^{2} - 1)

\int_{- 2}^{4} (\frac{x}{2} + 3) d x

What is the solution for y^{''}+8y^'+25y=0,y(0)=3,y^'(0)=-11 ?
The solution for y^{''}+8y^'+25y=0,y(0)=3,y^'(0)=-11 is y=e^{-4t}(3cos(3t)+1/3 sin(3t))