What is the solution for x^{''}-6x^'+9x=e^{3t} ?

The solution for x^{''}-6x^'+9x=e^{3t} is x=c_{1}e^{3t}+c_{2}te^{3t}+(e^{3t}t^2)/2

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x^{''}-6x^'+9x=e^{3t}

x^{^{''}} - 6 x^{^{'}} + 9 x = e^{3 t}

x = c_{1} e^{3 t} + c_{2} t e^{3 t} + \frac{e ^{3 t} t ^{2}}{2}

Solution steps

x^{''} (t) - 6 x^{'} (t) + 9 x = e^{3 t}

Solve linear ODE:

x = c_{1} e^{3 t} + c_{2} t e^{3 t} + \frac{e ^{3 t} t ^{2}}{2}

x = c_{1} e^{3 t} + c_{2} t e^{3 t} + \frac{e ^{3 t} t ^{2}}{2}

y^{^{''}} + 2 y^{^{'}} + 5 y = e^{- x} cos (2 x)

y^{^{''}} + 100 y^{^{'}} + 2500 y = 30, y (0) = 0, y^{^{'}} (0) = 2

\frac{d ^{2} x}{d t ^{2}} + x = 6 cos (2 t), x (0) = 3

\frac{1}{2} y^{^{''}} + 2 y = 4 tan (2 t) - \frac{1}{2} e^{2 t}

y^{^{''}} + 36 y = 12 cos (6 x) + 18 sin (6 x)

What is the solution for x^{''}-6x^'+9x=e^{3t} ?
The solution for x^{''}-6x^'+9x=e^{3t} is x=c_{1}e^{3t}+c_{2}te^{3t}+(e^{3t}t^2)/2