What is the solution for y^{''}-4y^'+4y=te^{2t} ?

The solution for y^{''}-4y^'+4y=te^{2t} is y=c_{1}e^{2t}+c_{2}te^{2t}+(e^{2t}t^3)/6

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y^{''}-4y^'+4y=te^{2t}

y^{^{''}} - 4 y^{^{'}} + 4 y = t e^{2 t}

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{e ^{2 t} t ^{3}}{6}

Solution steps

y^{''} (t) - 4 y^{'} (t) + 4 y = t e^{2 t}

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{e ^{2 t} t ^{3}}{6}

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{e ^{2 t} t ^{3}}{6}

\frac{d}{d x} (x^{\frac{2}{3}} - 3)

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n = 1 \sum \infty 3 x^{n}

\frac{4 x ^{2} - 7 x + 5}{2 x + 1}

What is the solution for y^{''}-4y^'+4y=te^{2t} ?
The solution for y^{''}-4y^'+4y=te^{2t} is y=c_{1}e^{2t}+c_{2}te^{2t}+(e^{2t}t^3)/6