What is the solution for 2x^{''}+5x^'+2x=10sin(t) ?

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} - 12 y = 3 e^{5} t

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

y^{^{''}} + 36 y = 2 cos (6 x) + 3 sin (6 x)

y^{^{''}} - 3 y^{^{'}} = t e^{3 t}, y (0) = 0, y^{^{'}} (0) = 2

y^{^{''}} - 12 y^{^{'}} + 35 y = - 40 e^{-} 3 t

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