What is the solution for y^{''}+(1/r)y^'=b ?

The solution for y^{''}+(1/r)y^'=b is y=c_{1}+c_{2}e^{-t/r}+rbt

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y^{''}+(1/r)y^'=b

y^{^{''}} + (\frac{1}{r}) y^{^{'}} = b

y = c_{1} + c_{2} e^{- \frac{t}{r}} + r b t

Solution steps

y^{''} (t) + (\frac{1}{r}) y^{'} (t) = b

Solve linear ODE:

y = c_{1} + c_{2} e^{- \frac{t}{r}} + r b t

y = c_{1} + c_{2} e^{- \frac{t}{r}} + r b t

y^{^{''}} - 10 y^{^{'}} + 61 = 0

\frac{d ^{2} x}{d t ^{2}} + 9 \frac{d x}{d t} + 8 x = 2 t

y^{^{''}} - 7 y^{^{'}} + 14 y = - 2 e^{3 t}, y (0) = - 6

y^{^{''}} - 5 y^{^{'}} + 6 y = x e^{3 x}

y^{^{''}} + 4 y = - cos (3 x) - 2 sin (3 x)

What is the solution for y^{''}+(1/r)y^'=b ?
The solution for y^{''}+(1/r)y^'=b is y=c_{1}+c_{2}e^{-t/r}+rbt