What is the solution for y^{''}+7y^'+12y=5x ?

The solution for y^{''}+7y^'+12y=5x is y=c_{1}e^{-3x}+c_{2}e^{-4x}+(5x)/(12)-35/144

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y^{''}+7y^'+12y=5x

y^{^{''}} + 7 y^{^{'}} + 12 y = 5 x

y = c_{1} e^{- 3 x} + c_{2} e^{- 4 x} + \frac{5 x}{12} - \frac{35}{144}

Solution steps

y^{''} (x) + 7 y^{'} (x) + 12 y = 5 x

Solve linear ODE:

y = c_{1} e^{- 3 x} + c_{2} e^{- 4 x} + \frac{5 x}{12} - \frac{35}{144}

y = c_{1} e^{- 3 x} + c_{2} e^{- 4 x} + \frac{5 x}{12} - \frac{35}{144}

\frac{d ^{2} y}{d x ^{2}} - 16 \frac{d y}{d x} - 64 y = 7 e^{8 x}

y^{^{''}} - 11 y^{^{'}} + 24 y = 2, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} + 2 y^{^{'}} + 10 y = 10 x + 2

y^{^{'''}} + 3 y^{^{''}} - 7 y^{^{'}} + 2 y = 3 x + 10 sin (x)

y^{^{''}} + 2 y^{^{'}} + y = (2 + 1) t e^{(- t)}

What is the solution for y^{''}+7y^'+12y=5x ?
The solution for y^{''}+7y^'+12y=5x is y=c_{1}e^{-3x}+c_{2}e^{-4x}+(5x)/(12)-35/144