(D^2 −4D+4)y = x^3 e^(2x)


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Question Number 103009 by bobhans last updated on 12/Jul/20

$(D^{2} - 4 D + 4) y = x^{3} e^{2 x}$
	Answered by bramlex last updated on 12/Jul/20

	$since y = e^{2 x} is a part of the$ $homogenous solution (from$ $solving the characteristic$ $equation χ^{2} - 4 χ + 4 = 0 =$ ${(χ - 2)}^{2} = 0 . we can use$ $reduction of order . set y = z . e^{2 x}$ $differentiating yields$ $Dy = (z^{'} + 2 z) e^{2 x} and D^{2} y =$ $(z^{″} + {4 z}^{'} + 4 z) e^{2 x} . then$ $substituting into the original$ $differential equation yields$ $e^{2 x} (z^{″} + {4 z}^{'} + 4 z) - {4 e}^{2 x} (z^{'} + 2 z) + 4 z . e^{2 x} = x^{3}$ $e^{2 x} which simplifies yields z^{″} = x^{3}$ $intgrating twice in succession$ $yields y = (C_{1} x + C_{2}) e^{2 x} + \frac{1}{20}$ $x^{5} e^{2 x} . hence a particular$ $solution is y = \frac{1}{20} x^{5} e^{2 x} . \oplus$
	Commented by bobhans last updated on 12/Jul/20

	$y e s . . .$
	Answered by mathmax by abdo last updated on 12/Jul/20

	$y^{^{″}} - {4 y}^{^{'}} + 4 = x^{3} e^{2 x}$ $(he) \to r^{2} - 4 r + 4 = 0 \Rightarrow {(r - 2)}^{2} = 0 \Rightarrow r = 2 \Rightarrow y_{h} = (ax + b) e^{2 x}$ $= {axe}^{2 x} + {be}^{2 x} = {au}_{1} + {bu}_{2}$ $W (u_{1}, u_{2}) = \| \begin{matrix} {xe}^{2 x} e^{2 x} \\ (2 x + 1) e^{2 x} {2 e}^{2 x} \end{matrix} \| = {2 xe}^{4 x} - (2 x + 1) e^{4 x} = - e^{4 x} \neq 0$ $w_{1} = \| \begin{matrix} o e^{2 x} \\ x^{3} e^{2 x} {2 e}^{2 x} \end{matrix} \| = - x^{3} e^{4 x}$ $W_{2} = \| \begin{matrix} {xe}^{2 x} 0 \\ (2 x + 1) e^{2 x} x^{3} e^{2 x} \end{matrix} \| = x^{4} e^{4 x}$ $v_{1} = \int \frac{W 1}{W} dx = \int \frac{- x^{3} e^{4 x}}{- e^{4 x}} = \int x^{3} dx = \frac{x^{4}}{4}$ $v_{2} = \int \frac{W_{2}}{W} dx = \int \frac{x^{4} e^{4 x}}{- e^{4 x}} = - \int x^{4} dx = - \frac{x^{5}}{5} \Rightarrow$ $y_{p} = u_{1} v_{1} + u_{2} v_{2} = {xe}^{2 x} \times \frac{x^{4}}{4} + e^{2 x} (- \frac{x^{5}}{5}) = \frac{x^{5}}{4} e^{2 x} - \frac{x^{5}}{5} e^{2 x}$ $= \frac{x^{5}}{20} e^{2 x} \Rightarrow the general solution is$ $y = y_{h} + y_{p} = (ax + b) e^{2 x} + \frac{x^{5}}{20} e^{2 x}$
	Commented by bemath last updated on 12/Jul/20

	$s i r A b d o & s i r B r a m l e x c o l l$
	Commented by mathmax by abdo last updated on 13/Jul/20

	$thanks sir$
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