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

The solution for y^{'''}+4y^{''}=3x-1 is y=c_{1}e^{-4x}+c_{2}+c_{3}x+(x^3)/8-(7x^2)/(32)

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

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

y = c_{1} e^{- 4 x} + c_{2} + c_{3} x + \frac{x ^{3}}{8} - \frac{7 x ^{2}}{32}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- 4 x} + c_{2} + c_{3} x + \frac{x ^{3}}{8} - \frac{7 x ^{2}}{32}

y = c_{1} e^{- 4 x} + c_{2} + c_{3} x + \frac{x ^{3}}{8} - \frac{7 x ^{2}}{32}

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What is the solution for y^{'''}+4y^{''}=3x-1 ?
The solution for y^{'''}+4y^{''}=3x-1 is y=c_{1}e^{-4x}+c_{2}+c_{3}x+(x^3)/8-(7x^2)/(32)