What is the solution for 2y^{''}+3y^'+y=gt ?

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

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2y^{''}+3y^'+y=gt

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 = 2 t, y (0) = - 2 t

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

y^{^{''}} + 2 y^{^{'}} + 5 y = 3 sin (t)

y^{^{''}} - y^{^{'}} - 6 y = 62 cos (3 x)

\frac{d ^{2} x}{d t ^{2}} + k \frac{d x}{d t} + g = 0

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