What is the solution for x^{'''}-4x^{''}+5x^'-2x=2t+3 ?

The solution for x^{'''}-4x^{''}+5x^'-2x=2t+3 is x=c_{1}e^{2t}+c_{2}e^t+c_{3}te^t-t-4

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x^{'''}-4x^{''}+5x^'-2x=2t+3

x^{^{'''}} - 4 x^{^{''}} + 5 x^{^{'}} - 2 x = 2 t + 3

x = c_{1} e^{2 t} + c_{2} e^{t} + c_{3} t e^{t} - t - 4

Solution steps

x^{'''} (t) - 4 x^{''} (t) + 5 x^{'} (t) - 2 x = 2 t + 3

Solve linear ODE:

x = c_{1} e^{2 t} + c_{2} e^{t} + c_{3} t e^{t} - t - 4

x = c_{1} e^{2 t} + c_{2} e^{t} + c_{3} t e^{t} - t - 4

y^{^{''}} + 4 y = sin (2 x + 3)

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

y^{^{''}} + y^{^{'}} - 12 y = - 40 t + 24 t^{2}

y^{^{''}} + y^{^{'}} + 4 y = e^{t} + e^{- t}

(d^{2} + 2 d + 2) y = 3 e^{x} sin (x)

What is the solution for x^{'''}-4x^{''}+5x^'-2x=2t+3 ?
The solution for x^{'''}-4x^{''}+5x^'-2x=2t+3 is x=c_{1}e^{2t}+c_{2}e^t+c_{3}te^t-t-4