What is the solution for y^{''}+6y^'+25y=0,y(0)=2,y^'(0)=14 ?

The solution for y^{''}+6y^'+25y=0,y(0)=2,y^'(0)=14 is y=e^{-3t}(2cos(4t)+5sin(4t))

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y^{''}+6y^'+25y=0,y(0)=2,y^'(0)=14

y^{^{''}} + 6 y^{^{'}} + 25 y = 0, y (0) = 2, y^{^{'}} (0) = 14

y = e^{- 3 t} (2 cos (4 t) + 5 sin (4 t))

Solution steps

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

Solve linear ODE:

y = e^{- 3 t} (2 cos (4 t) + 5 sin (4 t))

y = e^{- 3 t} (2 cos (4 t) + 5 sin (4 t))

\int sin (X) d X

x \to 2 + lim (\frac{- 1}{x ^{2} - 4})

\frac{\partial}{\partial x} (x^{2} + y^{2} + 2)

\int (e^{x}) d x

\frac{d}{d x} (\frac{2 tanh ( x )}{2})

What is the solution for y^{''}+6y^'+25y=0,y(0)=2,y^'(0)=14 ?
The solution for y^{''}+6y^'+25y=0,y(0)=2,y^'(0)=14 is y=e^{-3t}(2cos(4t)+5sin(4t))