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

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

y = \frac{1}{5 + 2 x}, (- 1, \frac{1}{3})

x \frac{d y}{d x} = x^{2} - y^{2} + y

\int - t d t

x \to 3 lim (x (\frac{2}{x} + 5))

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