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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x \to 4 lim (\frac{x}{x ^{2} - 16})

x \to 1 lim (\frac{x ^{2} - 3}{x - 6})

\int sin (\frac{π x}{L}) d x

\frac{d}{d x} (\frac{1}{3} x^{2})

y^{3} - (6 x + 6) + 3 x y^{2} \frac{d y}{d x} = 0

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