y + 3y = 0, y(0) = 2, y (0) = 3

Transcription

1 MATH 3 HOMEWORK #3 PART A SOLUTIONS Problem 311 Find the solution of the given initial value problem Sketch the graph of the solution and describe its behavior as t increases y + 3y 0, y(0), y (0) 3 Solution The characteristic equation is 0 r + 3r r(r + 3) which has roots r 3, 0 Thus the general solution is y c 1 + c e 3t Since y 3c e 3t, the initial conditions imply y(0) c 1 + c 3 y (0) 3c Thus c 1 and c 1 1, so the solution to the IVP is y 1 e 3t Note that y 1 as t y t Problem 311 Find the solution of the given initial value problem Sketch the graph of the solution and describe its behavior as t increases y + y y 0, y(0) 0, y (0) 1 Solution The characteristic equation is 0 r + r which has roots Then the general solution is r 1 ± which has derivative y c 1 1 y c 1 e 1 t + c e 1+ t e 1 t c 1 e 1+ t

2 The initial conditions imply so 0 y(0) c 1 + c 1 y (0) c c 1 1 c c 1 + c 1 ( Then c 1 and c, so the solution to the IVP is ) y e 1 t + e 1+ t Since 1 + > 0, the solution tends to as t 10 y 8 6 Problem 31 Solve the initial value problem t y y 0, y(0), y (0) β Then find β so that the solution approaches zero as t Solution The characteristic equation is 0 r 1, which has roots r ±1/ Then the general solution is y c 1 e t/ + c e t/ which has derivative y c 1 e t/ + c et/ The initial conditions imply y(0) c 1 + c β c 1 + c

3 The second equation is equivalent to β c 1 + c, so + β c, hence c 1 + β and c 1 1 β Thus the solution to the IVP is y (1 β)e t/ + (1 + β)e t/ The term e t/ tends to as t, so we set β 1 to eliminate it For β 1, we have y e t/ 0 as t Problem 316 Consider the initial value problem (see Example 5) where β > 0 y + 5y + 6y 0, y(0), y (0) β, (a) Solve the initial value problem (b) Determine the coordinates t m and y m of the maximum point of the solution as functions of β (c) Determine the smallest value of β for which y m (d) Determine the behavior of t m and y m as β Solution (a) The characteristic equation is 0 r + 5r + 6 (r + )(r + 3), which has roots r, 3 The general solution is with derivative The initial conditions imply y c 1 e t + c e 3t y c 1 e t 3c e 3t c 1 + c β c 1 3c so + β c Then c β so c β, so the solution to the IVP is (b) Differentiating, we have y (6 + β)e t + ( β)e 3t y (6 + β)e t 3( β)e 3t (6 + β)e t + 3( + β)e 3t Setting y 0 yields (6 + β)e t m 3( + β)e 3t m e t m Then y m 6 + β ) + β ( 3(+β) ( 3(+β) 8(6 + β)3 ( + β) 7( + β 3( + β) (6 + β) t m ln 3(+β) ( 3(+β) (6 + β)3 7( + β) 3 (6 + β) + β ( 3(+β) ( ( + β) (6 + β)

4 To ensure that this is a maximum, we compute y (t m ): y (t m ) 6 + β 9 9 (+β) (6+β) + β 7 (+β 8 (6+β 16 9 (6 + β ( + β) 8 (6 + β 3 ( + β) 8 (6 + β 9 ( + β) < 0 since β > 0 Thus (t m, y m ) is a maximum by the second derivative test (c) Observe that (6 + β 7( + β) (6 + β)3 7( + β) 0 (6 + β 7( + β) 0 β 3 9β 108β 16 0 (β + 3)(β (6 6 3))(β ( )) The only positive root of the above equation is β , so y m for β (d) As β, t m ln y m ( ( + β) ln (6 + β) (6 + β)3 7( + β) ( ) 3(/β + 1) ln(3/) (6/β + 1) Problem 39 Determine the longest interval on which the given initial value problem is certain to have a unique twice-differentiable solution Do not attempt to find the solution t(t )y + 3ty + y, y(3) 3, y (3) 1 Solution Rewriting the equation as y + 3 t y + t(t ) y t(t ) we see that the coefficient functions are discontinuous at t 0 and t Since t 0 3 is in (0, ), then (0, ) is the largest such interval Problem 310 Determine the longest interval on which the given initial value problem is certain to have a unique twice-differentiable solution Do not attempt to find the solution y + (cos(t))y + 3(ln t )y 0, y() 3, y () 1 Solution Since ln t is discontinuous at t 0 and t 0 > 0, then the desired interval is (0, ) Problem 38 Consider the equation y y y 0 (a) Show that y 1 (t) e t and y (t) e t form a fundamental set of solutions (b) Let y 3 (t) e t, y (t) y 1 (t) + y (t), and y 5 (t) y 1 (t) y 3 (t) Are y 3 (t), y (t), and y 5 (t) also solutions of the given differential equation? (c) Determine whether each of the following pairs forms a fundamental set of solutions: [y 1 (t), y 3 (t)]; [y (t), y 3 (t)]; [y 1 (t), y (t)]; [y (t), y 5 (t)]

5 Solution (a) The characteristic equation is 0 r r (r )(r + 1), which has roots r, 1 Thus e t and e t are both solutions Alternatively, one can show they are solutions simply by substituting them into the differential equation Computing the Wronskian, we find W(y 1, y )(t) 3e t which is nonzero for all t Thus y 1, y form a fundamental set (b) Yes, since they are all linear combinations of the fundamental solutions and the equation is homogeneous (c) We compute Problem 338 W(y 1, y 3 )(t) 6e t 0 for all t y 1, y 3 form a fundamental set W(y, y 3 )(t) 0 y, y 3 not a fundamental set W(y 1, y )(t) 6e t y 1, y form a fundamental set W(y, y 5 )(t) 0 y, y 5 not a fundamental set Solution Abel s formula, which we did not cover in class, is needed to solve this problem Please give all students full marks on this problem Problem 339 Solution Abel s formula, which we did not cover in class, is needed to solve this problem Please give all students full marks on this problem 5