What is the solution for x^{''}+3x^'+2x=sin(t) ?

The solution for x^{''}+3x^'+2x=sin(t) is x=c_{1}e^{-t}+c_{2}e^{-2t}+1/10 sin(t)-3/10 cos(t)

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x^{''}+3x^'+2x=sin(t)

x^{^{''}} + 3 x^{^{'}} + 2 x = sin (t)

x = c_{1} e^{- t} + c_{2} e^{- 2 t} + \frac{1}{10} sin (t) - \frac{3}{10} cos (t)

Solution steps

x^{''} (t) + 3 x^{'} (t) + 2 x = sin (t)

Solve linear ODE:

x = c_{1} e^{- t} + c_{2} e^{- 2 t} + \frac{1}{10} sin (t) - \frac{3}{10} cos (t)

x = c_{1} e^{- t} + c_{2} e^{- 2 t} + \frac{1}{10} sin (t) - \frac{3}{10} cos (t)

y^{^{''}} + 4 y = 2 sec (2 t)

\frac{d ^{2} y}{d x ^{2}} + 4 y = x sin (2 x)

2 y^{^{''}} - 7 y^{^{'}} + 5 y = - 29

y^{^{''}} + y = t^{2} + 2, y (0) = 1, y^{^{'}} (0) = - 1

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

What is the solution for x^{''}+3x^'+2x=sin(t) ?
The solution for x^{''}+3x^'+2x=sin(t) is x=c_{1}e^{-t}+c_{2}e^{-2t}+1/10 sin(t)-3/10 cos(t)