What is the solution for 4y^{'''}-y^{''}=3x ?

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

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4y^{'''}-y^{''}=3x

4 y^{^{'''}} - y^{^{''}} = 3 x

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

Solution steps

4 y^{'''} (x) - y^{''} (x) = 3 x

Solve linear ODE:

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

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

y^{^{''}} + 16 y^{^{'}} + 64 y = (36 x^{2} - 24 x + 12) e^{- 8 x}

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{t}, y (0) = 1, y^{^{'}} (0) = - 1

y^{^{''}} + 5 y^{^{'}} - 6 y = x e^{(- 5 x)}

y^{^{''}} + 4 y = cos (t)

y^{^{'''}} + 5 y^{^{''}} + 6 y^{^{'}} = 4 + 9 sin (x)

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