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

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

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

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

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

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

\int 1 + x^{2} x d x

\frac{d}{d x} (\frac{4 x + 9}{x})

\int \frac{x - 1}{x + 1} d x

\frac{d}{d x} (2 (x^{2} + 5 x - 2)^{- \frac{5}{7}})

\frac{d ^{2} y}{d t ^{2}} + 9 y = 0

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