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

The solution for y^{'''}+y^'=2 is y=c_{1}+c_{2}cos(t)+c_{3}sin(t)+2t

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

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

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

Solution steps

y^{'''} (t) + y^{'} (t) = 2

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} + 4 y = - 7 x + 4 x

\frac{d ^{2} y}{d x ^{2}} + a \frac{d y}{d x} = g

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

9 y^{^{''}} - 16 y = sin (x)

y^{^{'''}} + y^{^{'}} - 10 y = 29 e^{4 x}

What is the solution for y^{'''}+y^'=2 ?
The solution for y^{'''}+y^'=2 is y=c_{1}+c_{2}cos(t)+c_{3}sin(t)+2t