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

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

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

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

y = c_{1} + c_{2} cos (2 x) + c_{3} sin (2 x) + \frac{3 x ^{2}}{8} - \frac{x}{4}

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} cos (2 x) + c_{3} sin (2 x) + \frac{3 x ^{2}}{8} - \frac{x}{4}

y = c_{1} + c_{2} cos (2 x) + c_{3} sin (2 x) + \frac{3 x ^{2}}{8} - \frac{x}{4}

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