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

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

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

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

y = c_{1} e^{2 t} + c_{2} t e^{2 t} - e^{2 t} t + e^{2 t} t ln (t)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} t e^{2 t} - e^{2 t} t + e^{2 t} t ln (t)

y = c_{1} e^{2 t} + c_{2} t e^{2 t} - e^{2 t} t + e^{2 t} t ln (t)

y^{^{''}} - 5 y^{^{'}} + 7 y = 3 e^{5 t}

y^{^{''}} + y = sin (2 t), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 4 y^{^{'}} + 5 y = e^{2 x} tan^{2} (x)

y^{^{''}} - 2 y^{^{'}} + 5 y = 2 e^{- x}, y (0) = 0, y^{^{'}} (0) = 2

y^{^{''}} + 4 y^{^{'}} + 4 y = (x + 7) e^{- 2 x}

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