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

The solution for y^{''}-7y^'+12y=-e^x is y=c_{1}e^{4x}+c_{2}e^{3x}-1/6 e^x

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y^{''}-7y^'+12y=-e^x

y^{^{''}} - 7 y^{^{'}} + 12 y = - e^{x}

y = c_{1} e^{4 x} + c_{2} e^{3 x} - \frac{1}{6} e^{x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{3 x} - \frac{1}{6} e^{x}

y = c_{1} e^{4 x} + c_{2} e^{3 x} - \frac{1}{6} e^{x}

y^{^{''}} - 2 y^{^{'}} + y = e^{t} + t, y^{^{'}} (0) = 0, y (0) = 1

y^{^{''}} - y = 2 e^{x} - 1 - 6 \cdot x \cdot e^{x}

y^{^{'''}} + 4 y^{^{''}} - 5 y = x e^{x} + 17 x

x^{^{''}} - 4 x^{^{'}} + 4 x = 4 cos (2 t), x^{0} = 1, x^{^{'}} = - 1

y^{^{''}} + \frac{32}{3} y = cos (x)

What is the solution for y^{''}-7y^'+12y=-e^x ?
The solution for y^{''}-7y^'+12y=-e^x is y=c_{1}e^{4x}+c_{2}e^{3x}-1/6 e^x