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

The solution for y^{''}+2y^'+5y=0,y(0)=1,y^'(0)=2 is y=e^{-t}(cos(2t)+3/2 sin(2t))

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

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

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

Solution steps

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

Solve linear ODE:

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

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

n = 1 \sum \infty \frac{1}{n ^{3}}

\int 10 cos (100 x) d x

\frac{2}{x - 3}

\int \frac{1}{x ^{2} - 8 x + 65} d x

2 x + 1

What is the solution for y^{''}+2y^'+5y=0,y(0)=1,y^'(0)=2 ?
The solution for y^{''}+2y^'+5y=0,y(0)=1,y^'(0)=2 is y=e^{-t}(cos(2t)+3/2 sin(2t))