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**Unformatted text preview: **x = 2-1 3-2 ! x + 1-1 ! e t . First, note that x = 1 ! e t + 2 1 1 ! ( e t + te t ) = 3 2 ! e t + 2 1 1 ! te t . Now, on the other side of the diﬀerential equation we get: 2-1 3-2 ! x + 1-1 ! e t = 2-1 3-2 ! 1 ! e t + 2 1 1 ! te t ! + 1-1 ! e t = 2 · 1 + (-1) · 3 · 1 + (-2) · ! e t + 2 2 · 1 + (-1) · 1 3 · 1 + (-2) · 1 ! te t + 1-1 ! e t = 2 3 ! e t + 2 1 1 ! te t + 1-1 ! e t = 3 2 ! e t + 2 1 1 ! te t = x . So the given vector equation is a solution to the diﬀerential equation. 2...

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