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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 2 y = 0, y^{^{'}} (π) = 0, y^{^{'}} (0) = 1

y^{^{''}} + 2 y^{^{'}} + y = 0, y^{1} (0) = x e - x

y^{^{''}} - 4 y^{^{'}} - 5 y = 0, y (1) = 0

x^{^{''}} - 0.6 x^{^{'}} + 1.09 x = 0

((\frac{d ^{2} y}{d x ^{2}}) + 4 (\frac{d y}{d x}) + 20 y)^{2} = 0

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