What is the solution for x^{''}-ax=t ?

The solution for x^{''}-ax=t is x=c_{1}e^{sqrt(a)t}+c_{2}e^{-sqrt(a)t}-t/a

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x^{''}-ax=t

x^{^{''}} - a x = t

x = c_{1} e^{a t} + c_{2} e^{- a t} - \frac{t}{a}

Solution steps

x^{''} (t) - a x (t) = t

Solve linear ODE:

x = c_{1} e^{a t} + c_{2} e^{- a t} - \frac{t}{a}

x = c_{1} e^{a t} + c_{2} e^{- a t} - \frac{t}{a}

y^{^{''}} + 64 y = 16 sin (4 t), y (0) = \frac{1}{2}, y^{^{'}} (0) = 0

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

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

y^{^{''}} + 16 y = cos (4 t), y (0) = 0, y^{^{'}} (0) = 1

y^{^{'''}} + y^{^{'}} = 2 x^{2} + 4 sin (x)

What is the solution for x^{''}-ax=t ?
The solution for x^{''}-ax=t is x=c_{1}e^{sqrt(a)t}+c_{2}e^{-sqrt(a)t}-t/a