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

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

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

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

y = c_{1} + c_{2} cos (x) + c_{3} sin (x) + 2 x^{2}

Solution steps

y^{'''} (x) + y^{'} (x) = 4 x

Solve linear ODE:

y = c_{1} + c_{2} cos (x) + c_{3} sin (x) + 2 x^{2}

y = c_{1} + c_{2} cos (x) + c_{3} sin (x) + 2 x^{2}

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} = \frac{e ^{x}}{x ^{2} + 1}

\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} + 2 y = e^{3 x}

y^{^{''}} - \frac{25}{21} y^{^{'}} + \frac{5}{18} y = \frac{355}{21}

y^{^{''}} + 3 y^{^{'}} - 40 y = - 132 t + 80 t^{2}

y^{^{''}} + 8 y^{^{'}} + 12 = 0

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