What is the solution for y^{''}-4y^'-12y=te^{4t} ?

The solution for y^{''}-4y^'-12y=te^{4t} is y=c_{1}e^{6t}+c_{2}e^{-2t}-(e^{4t}t)/(12)-1/36 e^{4t}

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y^{''}-4y^'-12y=te^{4t}

y^{^{''}} - 4 y^{^{'}} - 12 y = t e^{4 t}

y = c_{1} e^{6 t} + c_{2} e^{- 2 t} - \frac{e ^{4 t} t}{12} - \frac{1}{36} e^{4 t}

Solution steps

y^{''} (t) - 4 y^{'} (t) - 12 y = t e^{4 t}

Solve linear ODE:

y = c_{1} e^{6 t} + c_{2} e^{- 2 t} - \frac{e ^{4 t} t}{12} - \frac{1}{36} e^{4 t}

y = c_{1} e^{6 t} + c_{2} e^{- 2 t} - \frac{e ^{4 t} t}{12} - \frac{1}{36} e^{4 t}

\int_{- 2}^{2} \frac{1}{2} (4^{2} - x^{4}) d x

\int_{0}^{1} 1 + 3 x d x

l a pl a ce t r an s f or m - 3 sin (3 t)

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

x \to 6 + lim (\frac{1}{x - 6})

What is the solution for y^{''}-4y^'-12y=te^{4t} ?
The solution for y^{''}-4y^'-12y=te^{4t} is y=c_{1}e^{6t}+c_{2}e^{-2t}-(e^{4t}t)/(12)-1/36 e^{4t}