What is the solution for linear y^{''}+2y^'+5y=0,y(0)=1,y^'(0)=1 ?

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

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linear y^{''}+2y^'+5y=0,y(0)=1,y^'(0)=1

linear

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 1, y^{^{'}} (0) = 1

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 18 y^{^{'}} + 162 y = 0

\frac{d ^{2} y}{d x ^{2}} - 8 \frac{d y}{d x} + 17 y = 0

x^{^{'''}} + 9 x^{^{''}} + 15 x^{^{'}} - 25 x = 0

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

a y^{^{''}} + b y^{^{'}} + cy = 0, y (0) = 0

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