Apps. and Interpretation for IBDP Maths Ebook 2 | Page 205

www . seprodstore . com 195 12 � Coupled differential equations : 1 . d d d d x ax by t y cx dy t � � � �� � � �� can be expressed as d d d d x a b x t y c d y t � � � � � �� � � � � � �� � � � � �� � � � � � 2 . 1 � , 2 � : Eigenvalues of a b c d � � � � � � 3 . 1 v , 2 v : Eigenvectors corresponding to 1 � and 2 � respectively 4 . 1 2 1 2 t t Ae Be � � � � x v v : Solution of the system 5 . Stable equilibrium if 1 � , 2 0 � � or 1 i a b � � � , 2 i a b � � � and 0 a � 6 . Unstable equilibrium if 1 � , 2 0 � � or 1 i a b � � � , 2 i a b � � � and 0 a � 7 . Saddle point if 1 2 0 �� � � Solving d d d d x ax by t y cx dy t � � � �� � � �� by Euler ’ s method , with 0 0 0 ( , , ) t x y and step length h : 1 n n t t h � and 1 ( , , ) 1 ( , , ) d d d d n n n n n n n n t x y n n t x y x x x h t y y y h t � � � � � � �� � � � � � Predator-prey models : d ( ) d d ( ) d x a by x t y cx d y t � � � �� � � �� , where a , b , c and d are positive constants � The second-order differential equation 2 2 d d 0 d d x x a bx t t � � � can be expressed as d d d d v av bx t x v t � � � � �� � �� Solutions of Chapter 12 SUMMARY POINTs

www . seprodstore . com 195 12 � Coupled differential equations : 1 . d d d d x ax by t y cx dy t � � � �� � � �� can be expressed as d d d d x a b x t y c d y t � � � � � �� � � � � � �� � � � � �� � � � � � 2 . 1 � , 2 � : Eigenvalues of a b c d � � � � � � 3 . 1 v , 2 v : Eigenvectors corresponding to 1 � and 2 � respectively 4 . 1 2 1 2 t t Ae Be � � � � x v v : Solution of the system 5 . Stable equilibrium if 1 � , 2 0 � � or 1 i a b � � � , 2 i a b � � � and 0 a � 6 . Unstable equilibrium if 1 � , 2 0 � � or 1 i a b � � � , 2 i a b � � � and 0 a � 7 . Saddle point if 1 2 0 �� � � Solving d d d d x ax by t y cx dy t � � � �� � � �� by Euler ’ s method , with 0 0 0 ( , , ) t x y and step length h : 1 n n t t h � and 1 ( , , ) 1 ( , , ) d d d d n n n n n n n n t x y n n t x y x x x h t y y y h t � � � � � � �� � � � � � Predator-prey models : d ( ) d d ( ) d x a by x t y cx d y t � � � �� � � �� , where a , b , c and d are positive constants � The second-order differential equation 2 2 d d 0 d d x x a bx t t � � � can be expressed as d d d d v av bx t x v t � � � � �� � �� Solutions of Chapter 12 SUMMARY POINTs