What is the solution for y^{''}-y^'-2y= 1/5 ?

The solution for y^{''}-y^'-2y= 1/5 is y=c_{1}e^{2t}+c_{2}e^{-t}-1/10

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y^{''}-y^'-2y= 1/5

y^{^{''}} - y^{^{'}} - 2 y = \frac{1}{5}

y = c_{1} e^{2 t} + c_{2} e^{- t} - \frac{1}{10}

Solution steps

y^{''} (t) - y^{'} (t) - 2 y = \frac{1}{5}

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- t} - \frac{1}{10}

y = c_{1} e^{2 t} + c_{2} e^{- t} - \frac{1}{10}

y^{^{''}} - 9 y = 5 e^{3 x}

y^{^{'''}} + 9 y^{^{''}} + 27 y^{^{'}} + 27 y = 3 e^{- 3 x}

y^{^{''}} + 16 y = 6 sin (4 x)

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

y^{^{'''}} - y^{^{''}} + y^{^{'}} - y = x e^{x} - e^{x} + 7

What is the solution for y^{''}-y^'-2y= 1/5 ?
The solution for y^{''}-y^'-2y= 1/5 is y=c_{1}e^{2t}+c_{2}e^{-t}-1/10