What is the solution for x^{''}-8x^'+16x=3te^{4t} ?

The solution for x^{''}-8x^'+16x=3te^{4t} is x=c_{1}e^{4t}+c_{2}te^{4t}+(e^{4t}t^3)/2

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x^{''}-8x^'+16x=3te^{4t}

x^{^{''}} - 8 x^{^{'}} + 16 x = 3 t e^{4 t}

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

Solution steps

x^{''} (t) - 8 x^{'} (t) + 16 x = 3 t e^{4 t}

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} + 3 y^{^{'}} + 2 = 0

(D^{2} - 2 D - 3) y = e^{3 x} + cos (x)

y^{^{''}} + 3 y^{^{'}} + 7 y = cos (t), y (0) = 0, y^{^{'}} (0) = 2

y^{^{''}} + 3 y^{^{'}} - 4 y = 2, y (0) = 0, y (0) = 0

4 y^{^{''}} + 4 y^{^{'}} + 1 = 0

What is the solution for x^{''}-8x^'+16x=3te^{4t} ?
The solution for x^{''}-8x^'+16x=3te^{4t} is x=c_{1}e^{4t}+c_{2}te^{4t}+(e^{4t}t^3)/2