What is the solution for y^{''}-7y^'+12y=-e^{4x} ?

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

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y^{''}-7y^'+12y=-e^{4x}

y^{^{''}} - 7 y^{^{'}} + 12 y = - e^{4 x}

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

Solution steps

y^{''} (x) - 7 y^{'} (x) + 12 y = - e^{4 x}

Solve linear ODE:

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

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

x^{^{''}} + x = \frac{1}{cos ( t )}

y^{^{''}} - 6 y^{^{'}} + 10 y = e^{3 x} cos (x)

y^{^{''}} + y = x^{2} + sin (x)

y^{^{''}} - 4 y^{^{'}} - 21 y = 320 e^{- t}

6 y^{^{''}} - y^{^{'}} - y = 50 e^{\frac{x}{2}} + 4 x^{2} - 5

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