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limemink388Lv1
6 Nov 2019
For the system of differential equations x' (t) = -9/5 x + 5/3 y + 2xy y' (t) = -18/5 x + 20/3 y = xy the critical point (x_0, y_0) with x_0 > 0, y_0 > 0 is x_0 Change variables in the system by letting x(t) = x_0 + u(t), y(t) = y_0 + v(t). The system for u, v is Use u and v for the two functions, rather than u(t)and v(t) For the u, v system, the Jacobian matrix at the origin is You should note that this matrix is the same as J (x_0, y_0) from the previous problem. Show transcribed image text For the system of differential equations x' (t) = -9/5 x + 5/3 y + 2xy y' (t) = -18/5 x + 20/3 y = xy the critical point (x_0, y_0) with x_0 > 0, y_0 > 0 is x_0 Change variables in the system by letting x(t) = x_0 + u(t), y(t) = y_0 + v(t). The system for u, v is Use u and v for the two functions, rather than u(t)and v(t) For the u, v system, the Jacobian matrix at the origin is You should note that this matrix is the same as J (x_0, y_0) from the previous problem.
For the system of differential equations x' (t) = -9/5 x + 5/3 y + 2xy y' (t) = -18/5 x + 20/3 y = xy the critical point (x_0, y_0) with x_0 > 0, y_0 > 0 is x_0 Change variables in the system by letting x(t) = x_0 + u(t), y(t) = y_0 + v(t). The system for u, v is Use u and v for the two functions, rather than u(t)and v(t) For the u, v system, the Jacobian matrix at the origin is You should note that this matrix is the same as J (x_0, y_0) from the previous problem.
Show transcribed image text For the system of differential equations x' (t) = -9/5 x + 5/3 y + 2xy y' (t) = -18/5 x + 20/3 y = xy the critical point (x_0, y_0) with x_0 > 0, y_0 > 0 is x_0 Change variables in the system by letting x(t) = x_0 + u(t), y(t) = y_0 + v(t). The system for u, v is Use u and v for the two functions, rather than u(t)and v(t) For the u, v system, the Jacobian matrix at the origin is You should note that this matrix is the same as J (x_0, y_0) from the previous problem.1
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Bunny GreenfelderLv2
4 Aug 2019