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

The solution for x^{''}+2x^'+5x=1,x(0)=0,x^'(0)=0 is x=e^{-t}(-1/5 cos(2t)-1/10 sin(2t))+1/5

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

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

x = e^{- t} (- \frac{1}{5} cos (2 t) - \frac{1}{10} sin (2 t)) + \frac{1}{5}

Solution steps

x^{''} (t) + 2 x^{'} (t) + 5 x = 1

Solve linear ODE:

x = e^{- t} (- \frac{1}{5} cos (2 t) - \frac{1}{10} sin (2 t)) + \frac{1}{5}

x = e^{- t} (- \frac{1}{5} cos (2 t) - \frac{1}{10} sin (2 t)) + \frac{1}{5}

y^{^{''}} + 9 y = e^{- 2 x}

y^{^{'''}} + 3 y^{^{''}} + 2 y^{^{'}} = x^{2} + 4 x + 8

y^{^{''}} + 4 y^{^{'}} + 4 y = - 2 sin (3 x)

x^{^{''}} - 2 x^{^{'}} + x = 2000 t^{2} - 150 e^{t}

y^{^{'''}} + 10 y^{^{''}} + 25 y^{^{'}} = e^{x}

What is the solution for x^{''}+2x^'+5x=1,x(0)=0,x^'(0)=0 ?
The solution for x^{''}+2x^'+5x=1,x(0)=0,x^'(0)=0 is x=e^{-t}(-1/5 cos(2t)-1/10 sin(2t))+1/5