APPLIED MATHEMATICS. Introduction to Differential Equations APM2A10. Supplementary Test 1: 23/03/2016

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1 APPLIED MATHEMATICS Introduction to Differential Equations APMA0 Supplementary Test : 3/03/06 Duration: 70 minutes Marks: 30 Assessor: Mr KD Anderson Moderator: Prof M Khumalo Instructions:. Answer all the questions.. All calculations must be shown. 3. All symbols have their usual meaning. 4. Pocket calculators are permitted. 5. All angles measured in radians. 6. No books or notes are allowed. Question (5 marks) Match each of the following differential equations to the corresponding direction field.. y = 8y 6y +y 3. y = xy 3. y = x y 4. y = xe y 5. y = 0.x +y Solution:. (e);. (b); 3. (a); 4. (d); 5. (c) Question (5 marks) For each of the differential equations, give the order of the differential equation and say whether it is linear or nonlinear. (a) x dx +4y = x3 x Solution: This is a first-order, linear differential equation. (b) ( x)y 4xy = cosx / 6

2 Solution: This is a first-order, linear differential equation. (c) x d y dx ( ) 4 +y = 0 dx Solution: This is a second-order, nonlinear differential equation. (d) d y dx = + ( ) dx Solution: This is a second-order, nonlinear differential equation. (e) (sinθ)y (cosθ)y = e y Solution: This is a third-order, nonlinear differential equation. Question 3 (7 marks) Find the solution of x dx +x(y +y ) = 0 and the largest interval on which the solution is defined. Solution: This differential equation can be solved by separation of variables or by noting that it is a Bernoulli equation. Separation of variables: We may rewrite the differential equation as x dx = x(y +y ) dx = (y +y ) y +y = dx The denominator of the fraction on the left-hand side can be simplified using the method of partial fractions, i.e. y +y = y(+y) = A y + B +y = A(+y) y(+y) + By y(+y) (A+B)y +A = A = y(+y) y +y = y (+y) and B = / 6

3 It follows that ( y +y ( y lny ln(+y) = ln ) +y = dx ) = x+c y +y = ke x, k = e c y(x) = k ex. Bernoulli equation: We may rewrite the differential equation as dx +y = y. ThisisaBernoulliequationwithn =. Wesubstitute u = y intoy +y = y, noting that y = u and dx = du dudx = u du dx, to obtain du u dx + u = u. Simplification yields du dx u = which is a linear differential equation in u. The integration factor is e dx = e x, and multiplying the previous equation by it yields e xdu dx e x u = d [ e x u ] = e x dx e x u = e x dx+c = e x +c and so u(x) = ce x = cex y(x) = ce x. Interval of definition: The largest interval on which the solution is defined is (, ). 3 / 6

4 Question 4 (7 marks) Find the solution of x dx +x(x+)y = ex, y() = 0. Solution: Rewriting the differential equation in standard form gives dx + x+ x y = x e x and note that it is a linear, first-order equation in y. Variation of parameters method: We are looking for u = u(x) such that y(x) = u(x)y (x) is the general solution to the differential equation, where y = y (x) is the solution to the complimentary equation dx + x+ x y = 0. The complimentary equation is a separable equation and it follows that dx = x+ x y y = dx lny = x lnx+c y = kx e x x dx+c Therefore y = x e x, and y = ux e x and y = u x e x + u( x 3 e x x e x ). Substituting y and y back into the differential equation gives ( u x e x +u( x 3 e x x e x ) ) + x+ x ( ux e x) = x e x u x e x = x e x u = e x u = e x dx+k and so u(x) = ex +k y(x) = x e x +kx e x. 4 / 6

5 Integration factor method: The integration factor is e x+ x dx = e + x dx = x e x. Multiplying the differential equation by the integration factor leads to x e x dx +x(x+)ex y = d [ x e x y ] = e x dx x e x y = e x dx+c x e x y = ex +c and so y(x) = x e x +cx e x. Particular solution: Applying the initial conditions y() = 0 yields e +c e = 0 e+ce = 0 c = e y(x) = ex e x x. Question 5 (6 marks) Consider the autonomous differential equation dx = y(y ). (a) Apply the existence and uniqueness theorem for nonlinear differential equations () to find an interval on which the solution to the autonomous differential equation will exist. Solution: By the existence and uniqueness theorem, both f and f/ y need tocontinuousonsomerectanglerforasolutiontoexist andbeunique. Note that f(y) = y(y ) and f y (y) = y y(y ) are only defined and continuous whenever y(y ) > 0, that is y < 0 and y >. Therefore an unique solution will exist for any (x 0,y 0 ) chosen in 5 / 6

6 either of the following two rectangles: R = { (x,y) R < x <, < y < 0 } R = { (x,y) R < x <, < y < } Therefore a solution exists for x R and y (,0) or y (, ). (b) Find the critical points of the autonomous differential equation and sketch the (4) phase line. Solution: Thecriticalpointsarefoundbysolvingf(y) = 0,i.e. y(y ) = 0. This implies that either y = 0 or y =. Therefore, the critical points are y = 0 and y =. Note that f(y) > 0 when y < 0 and y >. Thus, y is attracting for all y < 0 and y is repelling for all y >, nothing can be said about the behaviour of the fixed points for 0 < y <. The phase diagram is as follows: y 0 Information Theorem (Existence and uniqueness). (a) If f is continuous on an open rectangle R = { (x,y) R a < x < b, c < y < d } that contains (x 0,y 0 ), then the initial value problem dx = f(x,y), y(x 0) = y 0, has at least one solution on some open subinterval (a,b) that contains x 0. (b) If both f and f are continuous on R then the initial value problem y dx = f(x,y), y(x 0) = y 0, has a unique solution on some open subinterval of (a,b) that contains x 0. END OF QUESTION PAPER