What is the solution for y^{''}-4y^'+4y=5t^{-1}e^{2t} ?

The solution for y^{''}-4y^'+4y=5t^{-1}e^{2t} is y=c_{1}e^{2t}+c_{2}te^{2t}-5e^{2t}t+5e^{2t}tln(t)

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y^{''}-4y^'+4y=5t^{-1}e^{2t}

y^{^{''}} - 4 y^{^{'}} + 4 y = 5 t^{- 1} e^{2 t}

y = c_{1} e^{2 t} + c_{2} t e^{2 t} - 5 e^{2 t} t + 5 e^{2 t} t ln (t)

Solution steps

y^{''} (t) - 4 y^{'} (t) + 4 y = 5 t^{- 1} e^{2 t}

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} t e^{2 t} - 5 e^{2 t} t + 5 e^{2 t} t ln (t)

y = c_{1} e^{2 t} + c_{2} t e^{2 t} - 5 e^{2 t} t + 5 e^{2 t} t ln (t)

\frac{d}{d x} (x^{e x})

\frac{\partial}{\partial x} (x^{2} (x - y)^{- 2})

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

f (x) = (2 x^{2} + 1)

\int x^{2} sin (y) d y

What is the solution for y^{''}-4y^'+4y=5t^{-1}e^{2t} ?
The solution for y^{''}-4y^'+4y=5t^{-1}e^{2t} is y=c_{1}e^{2t}+c_{2}te^{2t}-5e^{2t}t+5e^{2t}tln(t)