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Question Number 127012 by Algoritm last updated on 26/Dec/20

	Answered by Olaf last updated on 26/Dec/20
	${ ((y_1 ′ = y_1 +y_2 + x (1))),((y_2 ′ = y_1 −2y_2 +2x (2))) :} (1) : y_2 = y_1 ′−y_1 −x (3) y_2 ′ = y_1 ′′−y_1 ′−1 (2) : y_1 ′′−y_1 ′−1 = y_1 −2(y_1 ′−y_1 −x)+2x y_1 ′′+y_1 ′−2y_1 = 1+4x ⇒ y_1 = ae^x +be^(−2x) −2x−(3/2) ⇒ y_1 ′ = ae^x −2be^(−2x) −2 (3) : y_2 = ae^x −2be^(−2x) −2−ae^x −be^(−2x) +2x+(3/2)−1 y_2 = −3be^(−2x) +2x−(3/2)$
	${\begin{cases} y_{1}^{'} = y_{1} + y_{2} + x (1) \\ y_{2}^{'} = y_{1} - 2 y_{2} + 2 x (2) \end{cases}$ $(1) : y_{2} = y_{1}^{'} - y_{1} - x (3)$ $y_{2}^{'} = y_{1}^{″} - y_{1}^{'} - 1$ $(2) : y_{1}^{″} - y_{1}^{'} - 1 = y_{1} - 2 (y_{1}^{'} - y_{1} - x) + 2 x$ $y_{1}^{″} + y_{1}^{'} - 2 y_{1} = 1 + 4 x$ $\Rightarrow y_{1} = {a e}^{x} + {b e}^{- 2 x} - 2 x - \frac{3}{2}$ $\Rightarrow y_{1}^{'} = {a e}^{x} - 2 {b e}^{- 2 x} - 2$ $(3) : y_{2} = {a e}^{x} - 2 {b e}^{- 2 x} - 2 - {a e}^{x} - {b e}^{- 2 x} + 2 x + \frac{3}{2} - 1$ $y_{2} = - 3 {b e}^{- 2 x} + 2 x - \frac{3}{2}$
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