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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{'''}} - 5 y^{^{''}} + 6 y^{^{'}} = 0 + 6 sin (x)

y^{^{''}} + y = 2 sin (2 t)

y^{^{''}} + y = sin (x), y^{^{'}} (\frac{π}{2}) = \frac{3}{2}, y (0) = 0

y^{^{''}} - 4 y^{^{'}} + 5 y = 125 t^{2}

\frac{d ^{3} y}{d x ^{3}} + \frac{d y}{d x} = tan (x)

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