What is the solution for y^{''}+6y^'+25y=0,y(0)=2,y^'(0)=-10 ?

The solution for y^{''}+6y^'+25y=0,y(0)=2,y^'(0)=-10 is y=e^{-3t}(2cos(4t)-sin(4t))

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y^{''}+6y^'+25y=0,y(0)=2,y^'(0)=-10

y^{^{''}} + 6 y^{^{'}} + 25 y = 0, y (0) = 2, y^{^{'}} (0) = - 10

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d x} (\frac{6 x}{1 + 3 x ^{2}})

x \to 0 lim (\frac{1}{x} ln (1 + x))

\int_{0}^{π} t^{2} e^{- s t} d t

y = e^{x}, y = e^{- x}, x = 1

x e^{- 5 x}

What is the solution for y^{''}+6y^'+25y=0,y(0)=2,y^'(0)=-10 ?
The solution for y^{''}+6y^'+25y=0,y(0)=2,y^'(0)=-10 is y=e^{-3t}(2cos(4t)-sin(4t))