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^{^{''}} + 4 y = 10 cos (2 x) + 15 sin (2 x)

y^{^{''}} - 5 y^{^{'}} = 3 cos (x)

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{2} x (cos (x) - 8 sin (x))

y^{^{''}} - 4 y = 6 cos (3 x + 3 0^{\circ})

y^{^{'''}} + y^{^{''}} + 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