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

The solution for y^{''}+2y^'+10y=0,y(0)=e^{-pi/6} is y=e^{-t}(1/(e^{pi/6)}cos(3t)+c_{2}sin(3t))

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y^{''}+2y^'+10y=0,y(0)=e^{-pi/6}

y^{^{''}} + 2 y^{^{'}} + 10 y = 0, y (0) = e^{- \frac{π}{6}}

y = e^{- t} (\frac{1}{e ^{\frac{π}{6}}} cos (3 t) + c_{2} sin (3 t))

Solution steps

y^{''} (t) + 2 y^{'} (t) + 10 y = 0

Solve linear ODE:

y = e^{- t} (\frac{1}{e ^{\frac{π}{6}}} cos (3 t) + c_{2} sin (3 t))

y = e^{- t} (\frac{1}{e ^{\frac{π}{6}}} cos (3 t) + c_{2} sin (3 t))

\frac{d ^{3} y}{d t ^{3}} - \frac{d ^{2} y}{d t ^{2}} = 0

y^{^{''}} + 4 y = 0, y (0) = 0, y^{^{'}} (0) = \frac{18}{5}

64 y^{^{''}} + 16 y^{^{'}} + 50 y = 0

x^{^{''}} - (nπ)^{2} x = 0

y^{^{''}} - 3 y^{^{'}} - 4 y = 0

What is the solution for y^{''}+2y^'+10y=0,y(0)=e^{-pi/6} ?
The solution for y^{''}+2y^'+10y=0,y(0)=e^{-pi/6} is y=e^{-t}(1/(e^{pi/6)}cos(3t)+c_{2}sin(3t))