What is the solution for y^{''}+4y^'+5y=0,y(0)=1 ?

The solution for y^{''}+4y^'+5y=0,y(0)=1 is y=e^{-2t}(cos(t)+c_{2}sin(t))

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

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

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

Solution steps

y^{''} (t) + 4 y^{'} (t) + 5 y = 0

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} (1) + 5 y = 0

\frac{d ^{2} x}{d t ^{2}} + 8 \frac{d x}{d t} + 16 x = 0

y^{^{'''}} + 2 y^{^{''}} - 10 y^{^{'}} + 4 y = 0

y^{^{'''}} - 2 y^{^{''}} - 4 y^{^{'}} = 0

y^{^{''}} - 14 y^{^{'}} + 65 y = 0, y (0) = 4, y^{^{'}} (0) = - 4

What is the solution for y^{''}+4y^'+5y=0,y(0)=1 ?
The solution for y^{''}+4y^'+5y=0,y(0)=1 is y=e^{-2t}(cos(t)+c_{2}sin(t))