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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{'''}} + y^{^{''}} + y^{^{'}} + y = e^{- t} + 11 t

y^{^{'''}} + y^{^{''}} + y^{^{'}} + y = e^{- t} + 12 t

y^{^{''}} + 3 y^{^{'}} = x^{2} e^{- 3 x}

y^{^{''}} + y = g

(D^{2} + 1) y = cos (x)

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