What is the solution for y^{''}+2y^'+5y=10 ?

The solution for y^{''}+2y^'+5y=10 is y=e^{-t}(c_{1}cos(2t)+c_{2}sin(2t))+2

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y^{''}+2y^'+5y=10

y^{^{''}} + 2 y^{^{'}} + 5 y = 10

y = e^{- t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + 2

Solution steps

y^{''} (t) + 2 y^{'} (t) + 5 y = 10

Solve linear ODE:

y = e^{- t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + 2

y = e^{- t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + 2

\frac{d}{d x} ((\frac{x + 1}{x - 1})^{2})

2 + x

f (t) = 6 t^{3} e^{- 2 t}

x \to 4 lim (\frac{x ^{2} - 9}{x - 3})

\frac{sec ^{2} ( x )}{2}, 2 cos^{2} (x)

What is the solution for y^{''}+2y^'+5y=10 ?
The solution for y^{''}+2y^'+5y=10 is y=e^{-t}(c_{1}cos(2t)+c_{2}sin(2t))+2