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Question Number 125651 by Algoritm last updated on 12/Dec/20

	Answered by Olaf last updated on 12/Dec/20
	${ ((y_1 ′ = y_1 +4y_2 (1))),((y_2 ′ = −y_1 +y_2 +e^(3x) (2))) :} (2) : y_1 = y_2 −y_2 ′+e^(3x) (3) ⇒ y_1 ′ = y_2 ′−y_2 ′′+3e^(3x) (1) y_2 ′−y_2 ′′+3e^(3x) = y_2 −y_2 ′+e^(3x) +4y_2 y_2 ′′−2y_2 ′+5y_2 = 2e^(3x) (E_H ) : y_(2H) ′′−2y_(2H) +5y_(2H) = 0 r^2 −2r+5 = 0 r = ((2±4i)/2) = 1±2i y_(2H) = e^x (Acos2x+Bsin2x) (E_0 ) : y_(2,0) ′′−2y_(2,0) +5y_(2,0) = e^(3x) y_(2,0) = αe^(3x) ⇒ 9α−2×3α+5α = 2 α = (1/4), y_(2,0) = (e^(3x) /4) y_2 = y_(2H) +y_(2,0) = e^x (Acos2x+Bsin2x)+(e^(3x) /4) ⇒ y_2 ′ = e^x [(A+2B)cos2x+(B−2A)sin2x]+((3e^(3x) )/4) (3) : y_1 = −2e^x (Bcos2x−Asin2x)−(e^(3x) /2)$
	${\begin{cases} y_{1}^{'} = y_{1} + 4 y_{2} (1) \\ y_{2}^{'} = - y_{1} + y_{2} + e^{3 x} (2) \end{cases}$ $(2) : y_{1} = y_{2} - y_{2}^{'} + e^{3 x} (3)$ $\Rightarrow y_{1}^{'} = y_{2}^{'} - y_{2}^{″} + 3 e^{3 x}$ $(1) y_{2}^{'} - y_{2}^{″} + 3 e^{3 x} = y_{2} - y_{2}^{'} + e^{3 x} + 4 y_{2}$ $y_{2}^{″} - 2 y_{2}^{'} + 5 y_{2} = 2 e^{3 x}$ $(E_{H}) : y_{2 H}^{″} - 2 y_{2 H} + 5 y_{2 H} = 0$ $r^{2} - 2 r + 5 = 0$ $r = \frac{2 \pm 4 i}{2} = 1 \pm 2 i$ $y_{2 H} = e^{x} (Acos 2 x + Bsin 2 x)$ $(E_{0}) : y_{2, 0}^{″} - 2 y_{2, 0} + 5 y_{2, 0} = e^{3 x}$ $y_{2, 0} = α e^{3 x}$ $\Rightarrow 9 α - 2 \times 3 α + 5 α = 2$ $α = \frac{1}{4}, y_{2, 0} = \frac{e^{3 x}}{4}$ $y_{2} = y_{2 H} + y_{2, 0} = e^{x} (Acos 2 x + Bsin 2 x) + \frac{e^{3 x}}{4}$ $\Rightarrow y_{2}^{'} = e^{x} [(A + 2 B) \cos 2 x + (B - 2 A) \sin 2 x] + \frac{3 e^{3 x}}{4}$ $(3) : y_{1} = - 2 e^{x} (Bcos 2 x - Asin 2 x) - \frac{e^{3 x}}{2}$
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