What is the solution for y^{'''}-4y^'=3x-1 ?

The solution for y^{'''}-4y^'=3x-1 is y=c_{1}+c_{2}e^{-2x}+c_{3}e^{2x}-(3x^2)/8+x/4

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y^{'''}-4y^'=3x-1

y^{^{'''}} - 4 y^{^{'}} = 3 x - 1

y = c_{1} + c_{2} e^{- 2 x} + c_{3} e^{2 x} - \frac{3 x ^{2}}{8} + \frac{x}{4}

Solution steps

y^{'''} (x) - 4 y^{'} (x) = 3 x - 1

Solve linear ODE:

y = c_{1} + c_{2} e^{- 2 x} + c_{3} e^{2 x} - \frac{3 x ^{2}}{8} + \frac{x}{4}

y = c_{1} + c_{2} e^{- 2 x} + c_{3} e^{2 x} - \frac{3 x ^{2}}{8} + \frac{x}{4}

y^{^{''}} - y^{^{'}} + 361 y = 19 sin (19 t)

y^{^{''}} - y^{^{'}} = 2 e^{x}

5 y^{^{''}} + 25 y^{^{'}} + \frac{1}{1 0 ^{- 2}} y = 10 e^{- \frac{0}{4}}

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

y^{^{''}} + 4 y^{^{'}} + 3 y = e^{t} + sin (t)

What is the solution for y^{'''}-4y^'=3x-1 ?
The solution for y^{'''}-4y^'=3x-1 is y=c_{1}+c_{2}e^{-2x}+c_{3}e^{2x}-(3x^2)/8+x/4