What is the solution for y^{''}+2y^'+5y=0,y(0)=0 ?

The solution for y^{''}+2y^'+5y=0,y(0)=0 is y=c_{2}e^{-t}sin(2t)

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y^{''}+2y^'+5y=0,y(0)=0

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 0

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

Solution steps

y^{''} (t) + 2 y^{'} (t) + 5 y = 0

Solve linear ODE:

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

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

x^{^{'''}} + x^{^{''}} - 12 x^{^{'}} = 0

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

y^{^{''}} - 5 y^{^{'}} + y = 0

\frac{d ^{2} x}{d t ^{2}} + 4 x = 0

x^{^{''}} + 16 x = 0, x (0) = 0, x^{^{'}} (0) = 5

What is the solution for y^{''}+2y^'+5y=0,y(0)=0 ?
The solution for y^{''}+2y^'+5y=0,y(0)=0 is y=c_{2}e^{-t}sin(2t)