What is the solution for u^{''}+au^'(t)=b ?

The solution for u^{''}+au^'(t)=b is u(t)=c_{1}+c_{2}e^{-at}+(bt)/a

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u^{''}+au^'(t)=b

u^{^{''}} + a u^{^{'}} (t) = b

u (t) = c_{1} + c_{2} e^{- a t} + \frac{b t}{a}

Solution steps

u^{''} (t) + a u^{'} (t) = b

Solve linear ODE:

u (t) = c_{1} + c_{2} e^{- a t} + \frac{b t}{a}

u (t) = c_{1} + c_{2} e^{- a t} + \frac{b t}{a}

6 y^{^{''}} + 3 y^{^{'}} - 30 y = 16 cos (x)

y^{^{''}} - 5 y^{^{'}} - 36 y = e^{9 x}

y^{^{''}} + y^{^{'}} + y = sin (x) + x + x^{2} - 1

y^{^{''}} + y = e^{2 x}, y (0) = 0

y^{^{''}} - 2 y^{^{'}} - 3 y = x e^{- x}

What is the solution for u^{''}+au^'(t)=b ?
The solution for u^{''}+au^'(t)=b is u(t)=c_{1}+c_{2}e^{-at}+(bt)/a