2A Differential Equations (a) For a differential equation of the form y = f(y), explain how f'(y) can be used to determine the stability of any equilibrium solutions and justify your answer. If f'(yo) > 0 then the equilibrium point yo is unstable, because locally az ~ f'(yo) (y - yo) has solution y ~ ef' (yo)x + yo, which diverges from yo. Similarly, if f'(yo) < 0 then the equilibrium point is stable. (b) Find the equilibrium solutions of the differential equation d.x dy = y3 - y2 - 2y and determine their stability. Sketch representative solution curves in the (x, y)-plane. [2] f(y) = y3 - y2 - 2y so f(y) = 0 at y = 0 or y = - 1 or y = 2. f'(y) = 3y2 - 2y - 2. So f'(0) = - 2 stable, f'(2) = 6 unstable [2] and f'(-1) = 3 unstable. Ay 2 × [3] [10] 7A Differential Equations (a) Define the Wronskian W of two solutions y1(x) and y2(x) of the differential equation y" +p(x)y' + q(x)y = 0 , and state a necessary and sufficient condition for y1(x) and y2(x) to be linearly independ- ent. Show that W (x) satisfies the differential equation (*) W'(x) =- p(x)W(x) . [2] W (x) = y1y2 - y'1y2 [2] Solutions are linearly independent if W(x) ¥ 0. W'(x) = y1/2+y1/2 -1231-12y" = y?(-p(x)y/2-q(x)y2)-y2(-p(x)y/- q(x)y1) =- p(x)W(x) [2]
(b) By evaluating the Wronskian, or otherwise, find functions p(x) and q(x) such that (*) has solutions y1(x) = 1 + cos x and y2(x) = sinx. [3] W(x)=(1+cosx) cosx - sinx(-sinx) = 1 + cosx. So p(x) =- W'(x)/W(x)=sinx/(1 + cosx). Then - sin x + 1+ cos x sin x cosx + q(x) sin x = 0 [3] What is the value of W (T)? Is there a unique solution to the differential equation for 0 6 x < oo with initial conditions y(0) = 0, y'(0) = 1? Why or why not? so q(x) =1- cosx/(1+ cosx) = 1/(1 + cos x) [1] W(T)=0 No. Using hint about W(T), both y = sinx and ( sifin + (fort cosac) for x >T [2] are solutions satisfying the initial conditions. (c) Write down a third-order differential equation with constant coefficients, such that y1(x) = 1+cos x and y2(x) = sin x are both solutions. Is the solution to this equation for 0 6 x < > with initial conditions y(0) = y"(0) = 0, y'(0) = 1 unique? Why or why not? [3] y'" + y'= 0. Solutions are y = 1, cosx and sinx. Since the Wronskian is now 1 sin x COS x 0 0 COS x - sin xl W = det! I =- 1/0 the solutions are linearty independentderall x. So the solution y = sin x obeys the initial conditions and is unique. [2] [20] 8A Differential Equations (a) The circumference y of an ellipse with semi-axes 1 and x is given by Z 2TT P 2