What is the solution for y^{''}+4y^'+5y=0,y(0)=-1,y^'(0)=-3 ?

The solution for y^{''}+4y^'+5y=0,y(0)=-1,y^'(0)=-3 is y=e^{-2t}(-cos(t)-5sin(t))

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y^{''}+4y^'+5y=0,y(0)=-1,y^'(0)=-3

y^{^{''}} + 4 y^{^{'}} + 5 y = 0, y (0) = - 1, y^{^{'}} (0) = - 3

y = e^{- 2 t} (- cos (t) - 5 sin (t))

Solution steps

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

Solve linear ODE:

y = e^{- 2 t} (- cos (t) - 5 sin (t))

y = e^{- 2 t} (- cos (t) - 5 sin (t))

\int sin (x) cos (n x) d x

\frac{d y}{d x} y = 8 x

x \to 0 + lim (\frac{x}{x - 3})

y^{^{''}} + 2 y^{^{'}} + 5 y = t^{3}

\frac{d}{d x} (\frac{x ^{2} + x - 2}{x ^{2}})

What is the solution for y^{''}+4y^'+5y=0,y(0)=-1,y^'(0)=-3 ?
The solution for y^{''}+4y^'+5y=0,y(0)=-1,y^'(0)=-3 is y=e^{-2t}(-cos(t)-5sin(t))