Answer Happy

Please explain only the yellow circled part. How did theycome up with the entries of the Jacobian?
Do not redo the question, only explain the entries for theJacobian in this particular question.
Question 6 second order non-linear differential equation: Classify the critical points of the plane autonomous system corresponding to the [10 marks] Solution Let x = y✓then ý = −x +v(1 − x²)y√. - For the critical points : * = 0 = y⇒x= x= 0 so x = 0. Therefore the critical point is (0, 0)✓. Jacobian is J J(0,0) (91) 7² − 4A = ₂² - 4 x+v(x² − 1)x + x = 0. - t = v ⇒ t > 0 (Unstable) and 7 < 0(Stable) A = 1 ν > 2: } = (v − 2)(v + 2)√ ⇒ −2 < v < 2 ⇒ 7² − 4 < 0 (Spiral) v>2 v < -2 Unstable node✓ 0 -1 – 2xyv v(1- 1 v (₁ ²2²)) ✓ ⇒ 7² − 4 > Onode
0 <v <2: Unstable Spiral✓ -2 <v < 0: stable Spiral✓ v < -2: stable node✓ v = 2: degenerate unstable node. v=-2: degenerate stable node.