What is the solution for y^{''}-4y^'+4y=t^{-3}e^{2t} ?

The solution for y^{''}-4y^'+4y=t^{-3}e^{2t} is y=c_{1}e^{2t}+c_{2}te^{2t}+(e^{2t})/(2t)

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y^{''}-4y^'+4y=t^{-3}e^{2t}

y^{^{''}} - 4 y^{^{'}} + 4 y = t^{- 3} e^{2 t}

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

Solution steps

y^{''} (t) - 4 y^{'} (t) + 4 y = t^{- 3} e^{2 t}

Solve linear ODE:

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

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

y^{^{''}} + 4 y^{^{'}} - 32 = 0

y^{^{''}} + 16 y = e^{5 t}, y (0) = 0, y^{^{'}} (0) = 0

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

y^{^{'''}} + 10 y^{^{''}} + 25 y^{^{'}} = e x

y^{^{''}} - y^{^{'}} + 2 y = x

What is the solution for y^{''}-4y^'+4y=t^{-3}e^{2t} ?
The solution for y^{''}-4y^'+4y=t^{-3}e^{2t} is y=c_{1}e^{2t}+c_{2}te^{2t}+(e^{2t})/(2t)