What is the solution for 2y^{''}-16y^'+32y=4e^{4t} ?

The solution for 2y^{''}-16y^'+32y=4e^{4t} is y=c_{1}e^{4t}+c_{2}te^{4t}+e^{4t}t^2

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2y^{''}-16y^'+32y=4e^{4t}

2 y^{^{''}} - 16 y^{^{'}} + 32 y = 4 e^{4 t}

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

Solution steps

2 y^{''} (t) - 16 y^{'} (t) + 32 y = 4 e^{4 t}

Solve linear ODE:

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

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

x \to 2 lim (x^{2} - 4 x - 2)

4 x^{2} + 3 x

\frac{\partial}{\partial x} (\frac{108}{x})

\int \frac{1}{e ^{3 t} + e ^{t}} d t

\frac{d y}{d x} = 4 x - 6

What is the solution for 2y^{''}-16y^'+32y=4e^{4t} ?
The solution for 2y^{''}-16y^'+32y=4e^{4t} is y=c_{1}e^{4t}+c_{2}te^{4t}+e^{4t}t^2