What is the solution for x^{''}+6x^'+5x=sin(t) ?

The solution for x^{''}+6x^'+5x=sin(t) is x=c_{1}e^{-t}+c_{2}e^{-5t}+1/13 sin(t)-3/26 cos(t)

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x^{''}+6x^'+5x=sin(t)

x^{^{''}} + 6 x^{^{'}} + 5 x = sin (t)

x = c_{1} e^{- t} + c_{2} e^{- 5 t} + \frac{1}{13} sin (t) - \frac{3}{26} cos (t)

Solution steps

x^{''} (t) + 6 x^{'} (t) + 5 x = sin (t)

Solve linear ODE:

x = c_{1} e^{- t} + c_{2} e^{- 5 t} + \frac{1}{13} sin (t) - \frac{3}{26} cos (t)

x = c_{1} e^{- t} + c_{2} e^{- 5 t} + \frac{1}{13} sin (t) - \frac{3}{26} cos (t)

2 \frac{d ^{2} y}{d x ^{2}} + 4 y^{^{'}} + 2 y = \frac{1}{e ^{x} ( x + 1 )}

y^{^{''}} + 4 y^{^{'}} + 4 y = 15 x + 17

y^{^{''}} + 16 y^{^{'}} + 64 y = 100 e^{2 x}

2 y^{^{''}} - 5 y^{^{'}} + 71.86 y = 4

y^{^{''}} - 2 y^{^{'}} - 3 y = 6 x e^{x} + 4 x - 5

What is the solution for x^{''}+6x^'+5x=sin(t) ?
The solution for x^{''}+6x^'+5x=sin(t) is x=c_{1}e^{-t}+c_{2}e^{-5t}+1/13 sin(t)-3/26 cos(t)