What is the solution for y^{'''}+y^'=2x^2 ?

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

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

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

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

Solution steps

y^{'''} (x) + y^{'} (x) = 2 x^{2}

Solve linear ODE:

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

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

f (x) = e^{t a n (x)}

\frac{\partial}{\partial u} (\frac{v}{1 + uv w})

g (t) = - 4 t^{2} + 3 t - 3

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{- t}

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