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

The solution for 2y^{''}+3y^'+y=g(t) is y=c_{1}e^{-t/2}+c_{2}e^{-t}+gt-3g

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2y^{''}+3y^'+y=g(t)

2 y^{^{''}} + 3 y^{^{'}} + y = g (t)

y = c_{1} e^{- \frac{t}{2}} + c_{2} e^{- t} + g t - 3 g

Solution steps

2 y^{''} (t) + 3 y^{'} (t) + y = g (t)

Solve linear ODE:

y = c_{1} e^{- \frac{t}{2}} + c_{2} e^{- t} + g t - 3 g

y = c_{1} e^{- \frac{t}{2}} + c_{2} e^{- t} + g t - 3 g

y^{^{''}} + y^{^{'}} = - 4 x + 2

y^{^{'''}} + 24 y^{^{''}} - 2048 y = e^{- 16 x}

(D^{2} - 1) y = (1 + e^{- x})^{- 2}

y^{^{''}} - 2 y^{^{'}} + y = 8 e^{t}

y^{^{''}} + y = 2 sin (2 t), y (0) = 6, y^{^{'}} (0) = 0

What is the solution for 2y^{''}+3y^'+y=g(t) ?
The solution for 2y^{''}+3y^'+y=g(t) is y=c_{1}e^{-t/2}+c_{2}e^{-t}+gt-3g