What is the solution for y^{''}+2y^'+17y=0,y(0)=3,y^'(0)=-2 ?

The solution for y^{''}+2y^'+17y=0,y(0)=3,y^'(0)=-2 is y=e^{-t}(3cos(4t)+1/4 sin(4t))

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y^{''}+2y^'+17y=0,y(0)=3,y^'(0)=-2

y^{^{''}} + 2 y^{^{'}} + 17 y = 0, y (0) = 3, y^{^{'}} (0) = - 2

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

Solution steps

y^{''} (t) + 2 y^{'} (t) + 17 y = 0

Solve linear ODE:

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

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

\int e^{- 4 x} sin (b x) d x

\int_{5}^{3} 2 d x

\frac{s + 2}{s ( s + 1 )}

\int 2 cot^{3} (x) csc^{3} (x) d x

x \to \infty lim (- 2 x^{4})

What is the solution for y^{''}+2y^'+17y=0,y(0)=3,y^'(0)=-2 ?
The solution for y^{''}+2y^'+17y=0,y(0)=3,y^'(0)=-2 is y=e^{-t}(3cos(4t)+1/4 sin(4t))