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Let $x = x_{1}(t), y = y_{1}(t)$ and $ x = x_{2}(t), y = y_{2}(t)$ be any two solutions of the linear non homogeneous system

$x' = p_{11}(t)x+p_{12}(t)y+g_{1}(t)$

$y' = p_{21}(t)x + p_{22}(t)+g_{2}(t)$

Show that $x = x_{1}(t) - x_{2}(t)$ and $ y = y_{1}(t) - y_{2}(t)$ is a solution of the corresponding homogeneous system.

Any help would be appreciated!

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    $\begingroup$ Just evaluate $x'(t)$ and $y'(t)$. $\endgroup$ – Math Lover Dec 12 '17 at 2:24
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It might be helpful to write the system this way: $$x' - p_{11}(t)x - p_{12}(t)y = g(t)$$ $$y' - p_{21}(t)x = p_{22}(t) + g(t)$$ So that the homogeneous system would be: $$x' - p_{11}(t)x - p_{12}(t)y = 0$$ $$y' - p_{21}(t)x = 0$$ Since $x_1, x_2, y_1, y_2$ solve the non-homogenous system, you have: $$x_1' - p_{11}(t)x_1 - p_{12}(t)y_1 = g(t)$$ $$y_1' - p_{21}(t)x_1 = p_{22}(t) + g(t)$$ and $$x_2' - p_{11}(t)x_2 - p_{12}(t)y_2 = g(t)$$ $$y_2' - p_{21}(t)x_2 = p_{22}(t) + g(t)$$ What happens when you subtract corresponding equations from each other?

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