ZJU UIUC Institute Prof Thomas Honold Spring Semester 2023 Homework 1 Di erential Equations Math 285 H1 We have considered the ODE y0 x y as an example in the lecture Actually there are four ODE s viz y0 x y and y0 y x which look very similar Draw direction elds for the other three ODE s and determine their solutions in both implicit and explicit form if possible H2 Determine all points t0 y0 2 R2 such that there is a unique solution on t0 1 of the IVP y0 p y y t0 y0 the value at time t0 determines the values at all future times t t0 H3 Let t0 y0 y1 2 R Show that the IVP has a unique solution y00 y y t0 y0 y0 t0 y1 H4 For each of the following ODE s determine at least one nonzero solution by using the Ansatz y x a e x or y x b x a y00 y2 b y00 5y0 6y 0 c y00 5y0 6y ex d y00 1 e 2x 1 y00 4x 2 y0 8y 0 f x2 1 x y00 2x 2 x y0 2 1 x y 0 2x y0 1 2x2 y 0 H5 Do two of the three exercises on the pendulum equation in BDM17 Ch 1 3 Exercises 23 25 in the 11th global edition Due on Fri Feb 24 4 pm Instructions For your homework it is best to maintain 2 notebooks which are handed in on alternate Fridays You may also use A4 sheets provided they are rmly stapled together Don t forget to write your name English and Chinese and your student ID on the rst page Homework is handed in on Fridays before the informal discussion session starts late homework won t be accepted and will be returned on the next Friday Answers to exercises must be justi ed it is not su cient to state only the nal result of a computation Answers must be written in English For a full homework score it is su cient to solve ca 80 of the mandatory homework exercises Optional exercises contribute to the homework score but they are usually more di cult and you should work on them only if you have su cient spare time Solutions prepared by TA s and TH 1 A y0 x y Rewriting the ODE as yy0 x 0 and integrating gives 1 2 y2 1 2 x2 C 2 R Replacing Figure 1 Direction eld of y0 x y C by C 2 turns this into y2 x2 C implicit form y x px2 C explicit form Explicit solutions have domain R if C 0 and domains 1 p C p C 1 if C 0 For C 0 the solutions viz y x x formally are not de ned at 0 since the domain of y0 x y excludes the x axis B y0 y x The solutions are y x Cx C 2 R with domains 1 0 0 1 Again the exclusion of x 0 is arti cial and due to the special form of the ODE It would be included if we rewrite the ODE as xy0 y 0 That all solutions have been found follows from y x 0 xy0 y x2 0 which implies y x C is a constant C y0 y x The solutions are y x C x C 2 R with domains 1 0 0 1 That these are 2 As discussed in the lecture the solutions of y0 p y form the 2 parameter family all solutions follows from xy 0 y xy0 0 which implies xy C is a constant y t 8 1 4 t c1 2 0 1 4 t c2 2 if t c1 if c1 t c2 if t c2 One or both of c1 c2 may be in nite c1 1 c2 1 Claim Solutions are uniquely determined for all t t0 i y0 y t0 0 If y0 0 then near t0 the solution must have the form y t 1 4 t c2 2 with c2 determined from 4 t0 c2 2 y0 i e t0 c2 2py0 and c2 t0 2py0 Note that c2 must be smaller than t0 in this 1 case Thus y t is determined for all t t0 2 Figure 2 Direction eld of y0 y x If y0 0 then y t is not uniquely determined since y1 t 0 for t t0 and y2 t 1 4 t t0 2 for t t0 are two di erent solutions If y0 0 then near t0 the solution has the form y t 1 4 t c1 2 form some c1 t0 determined as in the case y0 0 and hence y c1 0 Thus we are back in the previous case and the solution is not unique 3 If y t with domain R solves the given IVP then z t y t t0 solves the IVP z00 z z 0 y0 z0 0 y1 From Example 10 in the lecture we know that the unique solution of this IVP is z t y0 cos t y1 sin t Hence is unique as well y t z t t0 y0 cos t t0 y1 sin t t0 4 a y x bx y00 x b 1 x 2 b2x2 The only nonzero solution is 2 b 6 i e y x 6x 2 For this we use that b1x 1 b2x 2 holds for all x in an intervall of positive length i b1 b2 1 2 provided that both b1 b2 are nonzero The other Ansatz doesn t work because e x and e x 2 e2 x aren t scalar multiples of each other if 6 0 b The Ansatz y x b x doesn t work because in this case y00 5y0 6y involves x x 1 x 2 which don t cancel For y x ae x we obtain y00 5y0 6y 2ae x 5 ae x 6ae x 2 5 6 ae x which can be made zero by taking as a root of X 2 5X 6 i e 2 2 3 For a 1 this gives the two solutions y1 x e2x y2 x e3x Every linear combination y x a1 e2x a2 e3x a1 a2 2 R is then a solution as well 3 Figure 3 Direction eld of y0 y x c Again it is clear that y x b x doesn t work For y x ae x we see from b that we need to solve 2 5 6 ae x ex This can be done Set 1 and solve 2 5 6 a 2a 1 for a i e a 1 2 A solution is therefore y x 1 2 ex d Clearly only the 2nd Ansatz can work and in fact it does w l o g set b 1 y00 1 2x y0 1 2x2 y 1 x 2 2 X 1 if is a root of X 2 3 more generally y x a1x a2px a1 a2 2 R are solutions 2 i e …
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