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

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

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x^{''}-4x^'+3x=te^t

x^{^{''}} - 4 x^{^{'}} + 3 x = t e^{t}

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} + 3 y = 13 cos (2 x)

y^{^{'''}} - 6 y^{^{''}} + 12 y^{^{'}} - 8 y = t e^{3} + 1

y^{^{''}} + y^{^{'}} - 2 y = 2 (3 x + 1) e^{x} + 20 cos (2 x)

y^{^{''}} + 12 y = 8 e^{4 x}

y^{^{''}} + 2 y^{^{'}} + y = \frac{t}{e ^{t}}

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