What is the solution for y^{''}+2y^'+17y=0,y(0)=4,y^'(0)=-3 ?

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

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y^{''}+2y^'+17y=0,y(0)=4,y^'(0)=-3

y^{^{''}} + 2 y^{^{'}} + 17 y = 0, y (0) = 4, y^{^{'}} (0) = - 3

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

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

y^{^{''''}} + y^{^{'''}} = 0

(4 D^{4} + 20 D^{3} + 35 D^{2} + 25 D + 6) y = 0

(D^{2} - 4 D + 4) y = 0

25 x^{^{''}} - 20 x^{^{'}} + 4 x = 0

\frac{d ^{2} y}{d x ^{2}} = - k y

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