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

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

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x^{''}-2x^'+x=3te^{-t}

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

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

Solution steps

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

Solve linear ODE:

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

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

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