ODE Homework 2. Since M y N x, the equation is not exact. 2. Determine whether the following equation is exact. If it is exact, M y N x 1 x.

Transcription

1 ODE Homework.6. Exact Equations and Integrating Factors 1. Determine whether the foowing equation is exact. If it is exact, find the soution pe x sin qdx p3x e x sin qd 0 [.6 #8] So. Let Mpx, q e x sin, Npx, q e x sin 3x, then M e x cos, N x e x sin 3 Since M N x, the equation is not exact.. Determine whether the foowing equation is exact. If it is exact, find the soution 6x dx pn x qd 0, x 0 x [.6 #11] So. Let Mpx, q x 6x, Npx, q n x, then M N x 1 x which shows that the equation is exact. Thus there exists a function ψpx, q such that ψ x M, ψ N. Integrating Mpx, q with respect to x, we get» ψpx, q 6x dx gpq 3x n x gpq x for some function gpq. Now ψ n x g 1 pq n x, so gpq c 0 for some constant c 0. Hence the soution is given impicit as ψpx, q C, i.e., 3x n x C 3. Determine whether the foowing equation is exact. If it is exact, find the soution xdx px q 3 d 0 px q 3

2 [.6 #1] So. Let Mpx, q x px q 3 M N x, Npx, q 3x px q 5 px q 3, then which shows that the equation is exact. Thus there exists a function ψpx, q such that ψ x M, ψ N. Integrating Mpx, q with respect to x, we get» xdx 1 ψpx, q gpq a gpq px q 3 x for some function gpq. Now ψ px q 3 g 1 pq px q 3 so gpq c 0 for some constant c 0. Hence the soution is given impicit as ψpx, q C, i.e., x C 4. Sove the initia vaue probem p9x 1qdx p4 xqd 0, p1q 0 and determine at east approximate where the soution is vaid. [.6 #14] So. Let Mpx, q 9x 1, Npx, q x 4, then M N x 1 which shows that the equation is exact. Thus there is a function ψpx, q such that ψ x M, ψ N. Integrating Mpx, q with respect to x, we get» ψpx, q p9x 1qdx gpq 3x 3 x x gpq for some function gpq. Now ψ x g 1 pq x 4, so gpq c 0 for some constant c 0. Hence the soution is given impicit as ψpx, q C, i.e. 3x 3 x x C Since p1q 0, we have C 3 1, that is, the soution for the initia vaue probem is defined impicit b the equation 3x 3 x x,

3 In fact, according to the initia condition, can be expressed expicit as a function of x, which is x p3x 3 x q 0 pxq x a x 8p3x 3 x q x 4x 3 x 8x It is cear that the soution is defined for the region of x where 4x 3 x 8x Let P pxq 4x 3 x 8x 16, note that the discriminant for P pxq is which impies that P pxq 0 has on one rea root. B Soving P pxq 0, we have that x Since P p0q 16 0, we can concude that the soution for the initia vaue probem is vaid for x Show that the equation sin e x sin x dx cos e x cos x d 0 is not exact but becomes exact when mutipied the integrating factor µpx, q e x. [.6 #1] So. Let Mpx, q sin Then M e x sin x, Npx, q cos e x cos x. cos sin, N x e x cos x e x sin x which shows that the equation is not exact. Now mutipe the equation b µpx, q e x, we have that pe x sin sin xqdx pe x cos cos xqd 0 Note that pµmq pµnq x e x cos sin x which shows that the equation µmdx µnd 0 is exact. Thus there exists a function ψpx, q such that ψ x µm, ψ µn. Integrating µpx, qmpx, q with respect to x, we get» ψpx, q pe x sin sin xqdx gpq e x sin cos x gpq for some function gpq. Now ψ e x cos cos x g 1 pq e x cos cos x, so gpq c 0 for some constant c 0. Hence the soution is given impicit as ψpx, q C, i.e., e x sin cos x C

4 6. Show that if pn x M q{pxm Nq R, where R depends on the quantit x on, then the differentia equation M N 1 0 has an integrating factor of the form µpxq. formua for this integrating factor. [.6 #4] Find a genera Proof. Mutipe the origina equation b a function µ µpx, q. The equation µm µn 1 0 has in integrating factor if pµmq pµnq x, that is, µ M µ x N µn x µm µpn x M q Define R : Nx M xm N, then N x M RpxM Nq. If R is some function depending on on the quantit z : x. It foows that the modified form of the equation is exact, if µ M µ x N µrpxm Nq xµrm µrn This reation is satisfied if µ xµr and µ x µr Now consider µ µpzq µpxq. Then the partia derivatives are µ x µ 1 and µ xµ 1 with µ 1 dµ dz. Thus xµ 1 xµr and µ 1 µr which impies that µ must satisf µ 1 pzq µpzqrpzq. The ater equation is separabe, with dµ Rdz. Hence µ» µpzq exp Rpzqdz Therefore, given R Rpxq, it is possibe to determine µ µpxq which becomes an integrating factor of the differentia equation. 7. Find an integrating factor and sove the equation e x dx pe x cot csc qd 0 [.6 #9] So. Let Mpx, q e x, Npx, q e x cot M 0, N x e x cot csc. Then

5 which shows that the equation is not exact. However, since N x M M ex cot 0 cot e x which is on depend on, so the equation has an integrating factor µ µpq. Furthermore, we have that» µpq exp cot sds sin Mutipe the equation b µpq sin, we get the modified equation e x sin dx pe x cos qd 0 which is exact. Let Mpx, q e x sin, Npx, s q e x cos. Then there exists a function ψpx, q such that ψ x M, ψ N. s Integrating M with respect to x, we get» ψpx, q e x sin dx gpq e x sin gpq for some function gpq. Now ψ e x cos g 1 pq e x cos, so gpq c 0 for some constant c 0. Hence the soution is given impicit as ψpx, q C, i.e., e x sin C.7. Numerica Approximations: Euer s Method 8. Find approximate vaues of the soution of the given initia vaue probem 1 3 t, p0q 1 at t 0.1, 0., 0.3, and 0.4 using the Euer method with h 0.1. [.7 #1(a)] So. Let fpt, q 3 t. The Euer formua is of the form n 1 n f n h, n 0, 1,, where f n fpt n, n q 3 t n n. Since p0q 1, we set pt 0, 0 q p0, 1q. For h 0.1, we have that 1 0 p3 t 0 0 qh 1 0.1p3 0 1q 1. 1 p3 t 1 1 qh p q p3 t qh p q p3 t 3 3 qh p q

6 9. Find approximate vaues of the soution of the given initia vaue probem 1 3 cos t, p0q 0 at t 0.1, 0., 0.3, and 0.4 using the Euer method with h 0.1. [.7 #4(a)] So. Let fpt, q 3 cos t. The Euer formua is of the form n 1 n f n h, n 0, 1,, where f n fpt n, n q 3 cos t n n. Since p0q 0, we set pt 0, 0 q p0, 0q. For h 0.1, we have that 1 0 p3 cos t 0 0 qh 0 0.1p3 cos 0 0q p3 cos t 1 qh p3 cos q p3 cos t qh p3 cos q p3 cos t 3 3 qh p3 cos q Convergence of Euer s Method. It can be shown that, under suitabe conditions on f, the numerica approximation generated b the Euer method for the initia vaue probem 1 fpt, q, pt 0 q 0 converges to the exact soution as the step size h decreases. This is iustrated b the foowing exampe. Consider the initia vaue probem 1 1 t, pt 0 q 0. (a) Show that the exact soution is φptq p 0 t 0 qe t t 0 t. (b) Using the Euer formua, show that k p1 hq k 1 h ht k 1, k 1, (c) Noting that 1 p1 hqp 0 t 0 q t 1, show b induction that n p1 hq n p 0 t 0 q t n (i) for each positive integer n. (d) Consider a fixed point t t 0 and for a given n choose h t t 0 n n t for ever n. Note aso that h Ñ 0 as n Ñ 8. B substituting for h in Eq. (i) and etting n Ñ 8, show that n Ñ φptq as n Ñ 8. Hint: im p1 nñ8 a{nq n e a.

7 [.7 #0] Proof. (a) Note that the equation is 1 1 t which is inear with integrating is µptq exp ³ dt e t. Hence the soution φptq for the initia vaue probem is of the form φptq» t t 0 e t s p1 sqds 0 e t t 0 se t s st (b) Let fpt, q 1 t st 0 0 e t t 0 p 0 t 0 qe t t 0 t, then the Euer formua shows that k k 1 f k 1 h k 1 hfpt k 1, k 1 q k 1 hp1 t k 1 k 1 q p1 hq k 1 h ht k 1, k 1,, (c) The initia condition pt 0 q 0 automatica impies 1 p1 hq 0 h ht 0 p1 hqp 0 t 0 q t 0 h p1 hqp 0 t 0 q t 1 Assuming that the reation hods for some integer k 1, that is, k p1 hq k p 0 t 0 q t k Then b Euer formua, k 1 p1 hq k h ht k p1 hq p1 hq k p 0 t 0 q t k h ht k p1 hq k 1 p 0 t 0 q t k ht k h ht k p1 hq k 1 p 0 t 0 q t k h (d) p1 hq k 1 pt 0 t 0 q t k 1 Thus, b induction hpothesis, n p1 hq n p 0 t 0 q t n P N im n im nñ8 nñ8 im 1 nñ8 p 0 t 0 q im p1 hq n p 0 t 0 q t n t t np0 0 t 0 q t n t t n 0 1 t n t φptq nñ8 p 0 t 0 qe t t 0

8 3.1. Homogeneous Equations with Constant Coefficients 11. Find the genera soution of the differentia equation [ 3.1 #4] So. The characteristic equation is r 3r pr 1qpr q 0 Thus we get r 1,. Therefore the genera soution is of the form ptq c 1 e t c e t 1. Find the genera soution of the differentia equation [ 3.1 #7] So. The characteristic equation is r 9r 9 0 Thus we get r Therefore the genera soution is of the form ptq c 1 e p9 3 5qt c e p9 3 5qt t. Find the soution of the initia vaue probem , p0q 1, 1 p0q 0 Sketch the graph of the soution and describe its behavior as t increases. [ 3.1 #11] So. The characteristic equation is r 5r 3 0 Thus we get r 5. Therefore the genera soution is of the form ptq c 1 e 5 t c e 5 t

9 Thus 1 ptq p 5 qc 1 1, 1 p0q 0, so e 5 t c 1 c 1 5 c 1 5 p 5 qc e 5 c 0 t. Since p0q and we have that c 1 5, c 6 5. Hence the soution of the initia vaue probem 6 is ptq 5 6 Note that both 5 e 5 t 5 e 5 6 are negative, so e p 5 qt t Ñ 0 as t Ñ 8. Hence im tñ8 ptq 0. The graph of soution is as foows 14. Find the soution of the initia vaue probem , p1q 1, 1 p1q 0 Sketch the graph of the soution and describe its behavior as t increases. [ 3.1 #15] So. The characteristic equation is r 8r 9 0 Thus we get r 9, 1. Therefore the genera soution is of the form ptq c 1 e 9t c e t

10 Thus 1 ptq c e t 9c 1 e 9t. Since p1q 1, 1 p1q 0, so e 9 c 1 ec 1 9e 9 c 1 ec 0 and we have that c 1 e9, c Hence the soution of the 10e initia vaue probem is ptq e e 9t 10e et e9 9t 10 et 1 It is eas to see that e 9 9 im ptq im tñ8 tñ8 10 e 9t 10e et 9 10e im tñ8 et 8 The graph of soution is as foows 15. Sove the initia vaue probem 1 0, p0q α, 1 p0q. Then find α so that the soution approaches zero as t Ñ 8. [ 3.1 #1] So. The characteristic equation is r r 0 Thus we get r 1,. Therefore the genera soution is of the form ptq c 1 e t c e t Thus 1 ptq c e t c 1 e t. Since p0q α, 1 p0q, so c 1 c α c c 1

11 and we have that c 1 α, c 3 α. Hence the soution of 3 the initia vaue probem is ptq α e t α e t 3 3 In order to make ptq Ñ 0 as t Ñ 8, it suffice to set α such that the coefficient of e t to be zero. Thus, b etting α 3 0 ñ α, we have that ptq e t, and im ptq 0. tñ8 16. Consider the equation p3 αq 1 pα 1q 0 Determine the vaues of α, if an, for which a soutions tend to zero as t Ñ 8; aso determine the vaues of α, if an, for which a (nonzero) soutions become unbounded as t Ñ 8. [ 3.1 #4] So. The characteristic equation is r p3 αqr pα 1q 0 Thus we get r, α 1. Therefore the genera soution is of the form ptq c 1 e t c e pα 1qt It is cear that if α 1 0 ñ α 1, then e pα 1qt Ñ 0 as t Ñ 8. Thus, if α 1, then a soutions tend to zero as t Ñ 8. On the other hand, if α 1 0 ñ α 1, then e pα 1qt Ñ 8 as t Ñ 8. However, for the case c 0, then ptq c 1 e t for some nonzero constant c 1 and im ptq 0. Therefore, there is tñ8 no such vaue of α for which a soutions become unbounded.