What is the solution for -y^{''}+y^'+6y=1-x ?

The solution for -y^{''}+y^'+6y=1-x is y=c_{1}e^{-2x}+c_{2}e^{3x}-x/6+7/36

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-y^{''}+y^'+6y=1-x

- y^{^{''}} + y^{^{'}} + 6 y = 1 - x

y = c_{1} e^{- 2 x} + c_{2} e^{3 x} - \frac{x}{6} + \frac{7}{36}

Solution steps

- y^{''} (x) + y^{'} (x) + 6 y = 1 - x

Solve linear ODE:

y = c_{1} e^{- 2 x} + c_{2} e^{3 x} - \frac{x}{6} + \frac{7}{36}

y = c_{1} e^{- 2 x} + c_{2} e^{3 x} - \frac{x}{6} + \frac{7}{36}

y^{''} - 3 y^{'} + 2 y = 1 1^{(e^{3} x)}

y^{^{''}} - 2 y^{^{'}} + y = \frac{3 e ^{t}}{t ^{2} + 1}

y^{^{''}} - 36 y = 180 t - 72 e^{- 1}, y (0) = 0, y^{^{'}} (0) = 61

\frac{d ^{2} y}{d t ^{2}} + 2 \frac{d y}{d t} + 4 y = 13 t^{3}

y^{^{''}} + 2 y = - 4 x + 14

What is the solution for -y^{''}+y^'+6y=1-x ?
The solution for -y^{''}+y^'+6y=1-x is y=c_{1}e^{-2x}+c_{2}e^{3x}-x/6+7/36