What is the solution for y^{''}-4y^'+4y=6t^2e^{2t} ?

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

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y^{''}-4y^'+4y=6t^2e^{2t}

y^{^{''}} - 4 y^{^{'}} + 4 y = 6 t^{2} e^{2 t}

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

Solution steps

y^{''} (t) - 4 y^{'} (t) + 4 y = 6 t^{2} e^{2 t}

Solve linear ODE:

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

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

f (x) = ln (e x)

x \to - 3 lim (\frac{2 x + 8}{x + 3})

\frac{d}{d x} (\frac{x ^{4} - 12 x ^{2}}{( x ^{2} - 4 ) ^{2}})

\frac{\partial}{\partial x} (arcsin (\frac{a}{b} sin (x)))

\frac{48}{s ^{5}}

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