Exercises 2 - Elsevier
1Exercises 2.11. (a) yes; (b) no; (c) no3. f (x, y) y 1/5 , so fy (t, y) 15 y 4/5 is not continuous at (0, 0) so uniqueness 5/4is not guaranteed. Solutions:y 45 t,y 0( (p2 y, y 0y 1/2 , y 05. f (t, y) 2 y , so fy (t, y) is not 2 y, y 0 ( y)1/2 , y 0continuous at (0, 0). Therefore, the hypotheses of the Existence and UniquenessTheorem are not satisfied. 1dy3/2 y t dy t1/2 dt ln y 23 t3/2 C1 y Ce2t /3 .dty Application of the initial conditions yields y exp 23 (t3/2 1)9. Yes. f (t, y) sin y cos t and fy (t, y) cos y are continuous on a regioncontaining (π, 0).11. y sec t so y 0 sec t tan t y tan t and y(0) 1. fy (t, y) t is continuouson π/2 t π/2 and f (t, y) sec t is continuous on π/2 t π/2 so thelargest intervalpon which the solution 1is valid is π/2 t π/2.13. f (t, y) y 2 1 and fy (t, y) 2 (y 2 1) 1/2 ; unique solution guaranteedfor (a) only.7. Yes.15. (0, ). Solution is y 14 t 1 (t4 1).17. (0, ) because y ln t has domain t 019. ( , 1) because y 1/(t 1) has domain ( , 1) (1, ) and y 1/(t 3)has domain ( , 3) (3, )21. ( 2, 2)23. t 0, y t 1 sin t cos t, t 21.23.yy464322-6-42-2-21-4-2-1012t-646t
225. First, we solve the equation (see next section):dy y2dt1dy dty21 t Cy 1y .t CApplying the initial condition indicates that 1/C a C 1/a so a/(1 at). This solutions is defined for t 1/a or t -1.5-1.00.5-0.51.0t1.5-227. Separating variables (see next section) gives us y dy t dt 21 y 2 12 t2 C y C t2 . Applying the initial conditions indicates that C a2 soy a2 t2 (because y(0) a is positive). Thus, the interval of definition ofthe solution is t a26.27.y-2y22111-12t-21-1-1-1-2-22t
3Exercises 2.21. Separate variables and integrate:y 2 dy x dx1 31y x2 C3233y x2 C2 1/33 2y x C.21 2Cx C 2 x24x29 75. 3t 21 t4 52 y 14y C17. sinh 3x 2 cosh 4y C9. Separate variables and integrate:3. y 11dy dty 22t 11ln(2 y) ln(2t 1) C2 ln(2 y) ln 2t 1 C 2 y Celn 2t 1 2 y C 2t 1 y 2 C 2t 1. 11. y sin 1 34 cos x C13. Separate variables and integrate:dy kydt1dy k dtyln y kt Cy Ce ktIn the calculation above, remember that e kt C e kt eC . C is arbitrary so eCis positive and arbitrary. 1115. y cosh 1 120sinh 6t 16cosh 4t C17. y 31 ln 32 e2t C1119. 43 cos θ 12cos 3θ 16sin 4y sin y C221. y sin 2x 2xy 2y 5 4Cy 0 1123. 64 C cos (2 y (x)) 64 C cos (8 x 2 y) cos (8 x 2 y) 64 cos (2 y) 1
4 2 cos (8 x) 4 cos (4 x 2 y) 4 cos (4 x 2 y) 8 cos (4 x) 128 0 225. 41 x sin 2x 41 x 18 cos 2x 2 sin y C1127.58 6 sin (y) 2 cos (4 x) sin (y 4 x) sin ( y 4 x) 64 1 sin (y) 8 cos (2 x) 4 sin (y 2 x) 4 sin ( y 2 x) 64 C 64 C sin (y) 029. 32 (ln x)3/2 31 e3/y C31. Factor first, then separate, use partial fractions and simplify.dy/dt (t2 1)(y 2 1)1dyy2 1 111 dy2 y 1 y 1y 11ln2y 1y 1y 1 (t2 1) dt (t2 1)dt 1 2t t C32 2 Ce 3 t 2t2 2y 33. y 1 Ce 3 t1 Ce 2t2 23 t 2t13 12C 4x(3C 1)4C(x 1) 335.1dy dty3 1 12 y dy dt3(y 1) 3(y 2 y 1) 2y 1(y 1)1/3 3t 3 tan 1 3 ln C2(y y 1)1/63
5y-2yy2221111-12t-21-12t-21-1-1-1-1-2-2-2Figure 1: Direction fields for dy/dt y 3 y 2 , dy/dt y 3 y 2 , and dy/dt y 2 y 337.1dy dty3 y 1y 2dy dty y 1 1ln y ln y 2 1 t C2yp Cet2y 1Ce2ty2 1 Ce2trCe2ty 1 Ce2t39. 111 dy dt2(y 1) y 2(y 1)11ln y 1 ln y ln y 1 t C22py2 1 Cety1y2 1 Ce2tr1y 1 Ce2t2t
641. dy x3 dx y 41 x4 C. y(0) 14 · 04 C 0 C 0 so y (x) 1/4 x443. Integrating gives us x sin y C and applying the initial condition gives usC 2 so x (y) sin (y) -6-42-2 y (t) 2/3 9 3 t3/2y (t) 1 1 2 et 2 x, y (x) e 2 xy (x) ey (x) arctan (x) 1y 3 4(3x 1)1/3 y ln 12 e2x 12 e46y1, and for (c) it is57. The solution for (a) is y esin t , for (b) it is y 1 sin t 1y 4 r sin t sin2 t .4y1512.5107.552.5-7.5-5t-2.52.5-2.559. y 2x(x 2) 157.5
767. Use partial fractions.1dt dt12 4y y 2 11 dy 8 dty 2 y 6y 2 8t Clny 6y 2 Ce8ty 63Ce8t 1.y 2Ce8t 171. y exp c (t 1)t73. L(t) L e rB tExercises 2.31. The integrating factor is µ(t) e t . Multiplying through by the integratingfactor, applying the theorem, integrating and solving for y gives usdy y 10dte tdy ye t 10e tdtd t e y 10e tdte t y 10e t Cy 10 Cet .The preferred solution is to use undetermined coefficients. A general solution ofthe corresponding homogeneous equation, y 0 y 0 is yh Cet . The forcingfunction is f (t) 10. The associated set of functions for a constant is F {1}. Because no multiple of 1 is a solution to the corresponding homogeneousequation, we assume that a particular solution takes the form yp A · 1 A.Differentiating yp0 0 and substituting into the nonhomogeneous equation givesus yp0 yp A 10 so A 10 and yp 10. Therefore, a general solutionof the nonhomogeneous equation is y yh yp 10 Cet3. The integrating factor is µ(t) e t . Multiplying through by the integrating
8factor and integrating we havedy y 2 cos tdte tdy e t y 2e t cos tdtd t e y 2e t cos tdte t y e t cos t e t sin t Cy Cet cos t sin t.Observe that in integrating eat cos bt dt and eat sinbt dt either required integration1eat (a cos bt b sin bt)by parts twice or a table of integrals, eat cos bt dt 2a b21and eat sin bt dt 2eat ( b cos bt a sin bt).a b25. y Cet 2te t e t7. y 12 t Ct 1dy119. First write the equation in standard form y e x to see thatxxR dxp(x) 1/x. Then, an integrating factor is e 1/x dx xln x x. Multiplyingthrough by the integrating factor and integrating gives usdy y e xdxd(xy) e xdxxy e x Cxy Cx 1 x 1 e x . 11. y 4t2 1 C 4t2 1R13. An integrating factor is µ(t) e cot t dt e ln csc t sin t. Multiplyingthrough by the integrating factor, integrating the result, and solving for y resultsindy y cos t sin t cos tdtd(y sin t) sin t cos tdt1y sin t sin2 t C21y sin t C csc t,2Rwhich is equivalent to y 21 cos t cot t C csc t because sin t cos t dt 12 cos2 t C when choosing u cos t rather than u sin t when calculating the integral.sin t
915. An integrating factor is Zµ(t) exp 4t112.dt exp ln(4t 9) 4t2 924t2 9Multiplying through by the integrating factor, integrating the result, and solvingfor y results in dy4t11 y 4t2 9 dt4t2 9 4t2 9 d1 y dt4t2 91 y 4t2 9t4t2 9t 4t2 91p 24t 9 C4p1y (4t2 9) C 4t2 9.4 R17. An integrating factor is µ(t) e2 cot x dx e 2 ln csc x sin2 x. Multiplyingthe equation by the integrating factor, integrating and solving for y yieldssin2 x2dy 2y sin2 x cot x sin2 x cos xdx dsin2 x y sin2 x cos xdx1sin2 x y sin3 x C31y sin x C csc2 x.319. θ 1 Cer /221. x(y) 1 y Cey23. x(t) C/(t3 1)25. v(s) se s Ce s27. Use undetermined coefficients to find a general solution of the equation.yh Ce t . The associated set of functions for the forcing function f (t) e t isF {e t }. Because e t is a solution to the corresponding homogeneous equation,multiply F by tn where n is the smallest integer so that no element of tn F is asolution to the corresponding homogeneous equation. In this case, tF {te t }so we assume that a particular solution of the nonhomogeneous equation has theform yp Ate t . Differentiating yp , yp0 Ate t Ae t , and substituting intothe nonhomogeneous equation yields yp0 yp Ate t Ae t Ate t Ae t e t so A 1 and yp te t . Therefore a general solution of the nonhomogeneousequation is y yh yp Ce t te t . Application of the initial condition yields
10y1234t5!0.2y e t (t 1).!0.4!0.6!0.8!12R29. An integrating factor is µ(t) e 2t dt et . Multiplying the equation bythe integrating factor, integrating and solving for y yieldset222dy 2tet y 2tetdt2d t2 e y 2tetdt22et y et C2y 1 Ce t .y10.52Applying the initial condition yields y 1 2e t .12345!0.5!131. A general solution is y 2t 1 (t 1)et Ct 1 . Applying the initial conditionst
11y42yields y (2tet 2et 1)/t.246810t!2!4y42t2 1633. y 2t 4246810t!2!439. (a) y Cet 1; (b) y Ce t t; (c) y Ce t sin t; (d) y Cet e t43. y 0 y t has solution y t 1 Ce t . y(0) 1 C 2 so y t 1 2e tfor 0 t 1. When t 1, y 1 1( 2e 1 2/e. The solution to y 0 y 0,t 1 2e t , 0 t 1y(1) 2/e is y 2e t . Thus, y(t) .2e t , t 1(e 2t , 0 t 145. y(t) e2 4t , t 147. y (t) 2/5 cos (2 t) 1/5 sin (2 t) Cet49. y (t) (t C) e t51. y (t) 1/25 1/5 t Ce5 t53. y (t) 2 cos (t) 4 sin (t) 4 et Ce1/2 t55. y (t) 2/11 et Ce 10 t57. y (t) (2 t C) et59. y (t) cos (t) sin (t) t 1 Ce t63. y (t) t 1 Ce t , y (t) 1/2 cos (t) 1/2 sin (t) Ce t , y (t) 1/2 cos (t) 1/2 sin (t) Ce t , y (t) 1/2 et Ce t65. y (t) t 1 e t , y (t) t 1, y (t) t 1 e t , y (t) t 1 2 e t ,y (t) t 1 3 e t
12Exercises 2.41. My (t, y) 2y 12 t 1/2 Nt (t, y), exact3. My (t, y) cos ty ty sin ty Nt (t, y), exact5. The equations is exact because t (sty 2 ) 3y 2 y (y 3 )7. My (t, y) sin 2t 6 2 sin 2t Nt (t, y), not exact9. My (t, y) y 1 Nt (t, y), exact11. y C t313. y 0, ty 2 C (2t y 3 ) 3y 2 (3ty 2 4).15. Observe that the equation is exact because y t32LetR F (t,3y) have 2total 3derivative (2t y ) dt (3ty 4) dy. Then, F (t, y) (2t y ) dt t ty g(y). Differentiating F with respect to y, Fy (t, y) 3ty 2 g 0 (y) 3ty 2 4 g 0 (y) 4 so g(y) 4y and F (t, y) t2 ty 3 4y. Ageneral solution is then t2 ty 3 4y C or t2 ty 3 4 y (t) C. 217. The equation is exact because(2ty) 2t (t y 2 ). Let F (t, y) y tRsatisfy Ft (t, y)dt Fy (t, y)dy 2ty dt (t2 y 2 ) dy. Then, F (t, y) 2ty dt t2 y g 0 (y) t2 y 2 so g 0 (y) y 2 g(y) 31 y 3 . Therefore F (t, y) t2 13 y 3and a general solution of the equation is t2 31 y 3 C. Observe that solvingthis as a homogeneous equation of degree 2 (see the nextsection) results in the !y 3 t2 y 2 ln (t) C.following form of the solution: 1/3 lnt3 19. The equation is exact because(sin2 y) 2 sin y cos y sin 2y (t sin 2y). y t2LetF(t,y)satisfyF(t,y)dt F(t,y)dy sinydt tsin2ydy.Then,F (t, y) tyRsin2 y dt t sin2 y g(y) so g 0 (y) 0 g(y) 0 and F (t, y) t sin2 y. Ageneral solution is then t sin2 y C or ln t 2 ln sin y C. t(e sin y) et cos y (1 et cos y).21. The equation is exact because y tLet F (t, y)R satisfy Ft (t, y)dt Fy (t, y)dy et sin y dt (1 et cos y) dy. Then,F (t, y) et sin y dt et sin y g(y) so g 0 (y) 1 g(y) t. Thus, F (t, y) et sin y t and a general solution of the equation is et sin y y -2-4-4-6-646t-4 23. y 0, y C sec t2 tan t225. The equation is exact because (1 y 2 cos ty) 2y cos ty ty 2 sin ty y
13 (sin ty ty cos ty). Let F (t, y) satisfy Ft (t, y)dt Fy (t, y)dy (1 y 2 cos ty) dt tR(sin ty ty cos ty) dy. Then, F (t, y) (1 y 2 cos ty) dt t y sin ty g(y) soFy (t, y) sin ty ty cos yt g 0 (y) so g 0 (y) 0 and g(y) 0. Then F (t, y) t y sin(yt) and a general solution of the equation is t y sin yt C. ((3 t) cos(t y) sin(t y)) cos(t 27. The equation is exact because y y) (3 t) sin(t y) ((3 t) cos(t y)). Let F (t, y) satisfy Ft (t, y)dt tFy (t, y)dy R ((3 t) cos(t y) sin(t y)) dt (3 t) cos(t y) dy. Then,F (t, y) ((3 t) cos(t y) sin(t y)) dt (3 t) sin(t y) g(y) soFy (t, y) (3 t) cos(t y) g 0 (y) (3 t) cos(t y) g 0 (y) 0 g(y) 0so F (t, y) 3 sin(t y) t sin(t y) and 3 sin(t y) t sin(t y) C. t 2 y 2 ey/t 1 t 3 yey/t (2t y) 29. The equation is exact because y y/te (1 y/t) . Let F (t, y) satisfy Ft (t, y)dt Fy (t, y)dy t 2 y 2 ey/t 1 dt t ey/t (1 y/t) dy. Then, F (t, y) t 2 y 2 ey/t 1 dt t yey/t g(y). Next,Fy (t, y) ey/t (1 y/t) g 0 (y) ey/t (1 y/t) so g 0 (y) 0 g(y) 0. Thus,F (t, y) yey/t t and yey/t t -2-2-2-4-4-4-6-6-646t 2ty 2 4ty 2t2 y . Let F (t, y) y tRsatisfy Ft (t, y) dt Fy (t, y) dy 2ty 2 dt 2t2 y dy. Then, F (t, y) 2ty 2 dt t2 y 2 g(y) so Fy (t, y) 2t2 y g 0 (y) 2t2 y, which means that g 0 (y) 0 g(y) 0. Therefore F (t, y) t2 y 2 and a general solution (or, integral curves) ofthe differential equation are t2 y 2 C. Application of the initial condition resultsin y 2 t2 1 0.31. This equation is exact becauseObserve that dividing the differential equation by 2ty yields y dt t dy 0,dy1dy y 0 or y 0. This is a first order linearwhich is equivalent to tdtdttRhomogeneous equation with integrating factor µ(t) e 1/t dt t. Multiplying
14through by the integrating factor, integrating and solving for y gives ustdy y 0dtd(ty) 0dtty Cy Ct 1 .Observe that squaring both sides of the equation and solving for C gives us thesame result as that obtained by solving the equation as an exact equation. 2 2ty 3t2 2t t 1 . Let33. The equation is exact because y t 22F (t, y) satisfyF(t,y)dt F(t,y)dy 2ty 3tdt t 1 dy. Then,ty R2232F (t, y) 2ty 3t dt t y t g(y) so Fy (t, y) t g 0 (y) t2 10 g (y) 1 g(y) y. Therefore, F (t, y) t2 y t3 y and the integralcurves are t2 y t3 y C. Applying the initial condition and solving for y gives t3 1us y .(t 1) (t 44t y(e 2ty) ey 2t tey t2 . y t Let F (t, y)R satisfy Ft (t, y) dt Fy (t, y) dy (ey 2ty) dt tey t2 dy. Then,F (t, y) (ey 2ty) dt tey t2 y g(y). Differentiating with respect to y,Fy (t, y) tey t2 g 0 (y) tey t2 g 0 (y) 0 g(y) 0 so F (t, y) tey t2 yand the integral curves are tey t2 y C. Applying the initial condition resultsin tey t2 y 0. 37. The equation is exact becausey 2 2 sin 2t 2y (1 2ty). Let y t 2F (t, y) satisfyR F2 t (t, y) dt F (t, y) dy2 y 2 sin 2t dt (1 2ty) dy. Then,F (t, y) y 2 sin 2t dt ty cos 2t g(y). Differentiating with respect to y, Fy (t, y) 2ty g 0 (y) 1 2ty g 0 (y) 1 g(y) y soF (t, y) ty 2 cos 2t y and the integral curves are ty 2 cos 2t y C. Applying the initial condition results in ty 2 cos (2 t) y 2 0.35. The equation is exact because
44t 12 y 2y ( 2ty).21 t t 1Let F (t, y) satisfy Ft (t, y) dt Fy (t, y) dy y 2 dt 2ty dy. Then,21 t R12 ydt tan 1 t ty 2 g(y). Differentiating with reF (t, y) 1 t2spect to y, Fy (t, y) 2ty g 0 (y) 2ty g 0 (y) 0 g(y) 0 soF (t, y) tan 1 t ty 2 and the integral curves are tan 1 t ty 2 C. Aparctan (t)plying the initial condition results in y 2 0.t39.The equation is exact because y 38.39.y-6-4y6644222-246t-6-42-2-2-2-4-4-6-646t41. (a) The equation is exact because ( 2x y cos(xy)) cos(xy) xy sin(xy) (2y x cos(xy)) . y xThe integral curves for the solution take the form F (x, y) C, whereZF (x, y) ( 2x y cos(xy)) dx x2 sin(xy) g(y).Because F (x, y) x2 sin(xy) g(y) 2y x cos(xy) g 0 (y), y yg 0 (y) 2y so g(y) y 2 , F (x, y) y 2 x2 sin(xy), and the integral curves ofthe equation are given by y 2 x2 sin(xy) C. Applying the initial condition
16y(0) 0 results in 0 C so that the solution to the initial value problem isy 2 x2 sin(xy).Observe in the following figure that the initial condition y(0) 0 does not resultin a unique b) Writing the differential equation in differential form gives usdy2x y cos(xy) .dx2y x cos(xy){z} f (x,y)Using the notation in the theorem, x 4 x2 y 2 sin(xy) cos(2xy) 9 f . y2(x cos(xy) 2y)2The results do not contradict the Theorems because neither of these functions iscontinuous on a region containing the origin, (0, 0).43. Here,dx2xyy2 xdy p pand. Then,dtdtx2 y 2x2 y 2dy/dty2 xdy dxdx/dt2xy x y 2 dx 2xy dy 0.Using the notation in section, M (x, y) x y 2 so My (x, y) 2y and N (x, y) 2xy so Nx (x, y) 2y so the equation x y 2 dx 2xy dy 0 is exact. LetF (x, y) be the potential function. Then,Z F (x, y) x y 2 dx1F (x, y) x2 xy 2 g(y).2
17Next, Fy (x, y) 2xy g 0 (y) 2xy so g 0 (y) 0. We choose g(y) 0 so that11F (x, y) x2 xy 2 and the integral curves are given by x2 xy 2 /3(et t2 ) t C 0y 1 t2254yt t (y) 1/4 t4 Ccos (y (t)) t2 sin (y) t Ct2 y cos ty C159. (a) (i) y t, t2 2ty y 2 C (ii) y Ct C 112 2y 1920 Ct 20 39C t 40CHiL110 HiiLy-1.02-2-249. y 51.y 10C 2 t2 10 .51.0t-1.0-0.50.5-0.5-0.5-0.5-1.0-1.0-1.01.0tx
18Exercises 2.5dwdy1. This is Bernoulli with n 1. Let w y 1 ( 1) y 2 2ydtdtdy1 1 dw y. Then,dt2dtdy 1 ydt21 1 dw 1y y2dt2dw y2dtdw wdt ty 1 ty 1 2t 2t.Use undetermined coefficients to solve for w. A general solution of the corresponding homogeneous equation is wh Cet . The associated set of functions forthe forcing function f (t) 2t is F {t, 1}. Because no element of F is a solutionto the corresponding homogeneous equation we assume that a particular solutionhas the form wp At B wp0 A. Substituting wp into the nonhomogeneousequation yields wp0 wp At (A B) 2t so A 2 and B 2 sowp 2t 2. A general solution is then w wh wp Cet 2t 2. Becausew y 2 , y Cet 2t 2.dwdy3. This is Bernoulli with n 3 so we let w y 1 3 y 2 . Then, 2y 3dtdt1 3 dwdyso y . Substituting into the equation gives us2 dtdtdw y 2ty 3 cos tdtdw 1 2 y 2 cos tdttdw 1 w 2 cos tdttd(tw) 2t cos tdttw 2 cos t 2t sin t C ty 3w 2t 1 cos t 2 sin t Ct 11 2t 1 cos t 2 sin t Ct 1y21y . 2t 1 cos t 2 sin t Ct 1p (2 cos (t) 2 t sin (t) C) t2 cos (t) 2 t sin (t) C935. y 3/2 20cos (t) 20sin (t) Ce3 t 0or y
193tt3 3C9. This is a Bernoulli equation with n 2 so we let w y 1 2 y 1 1/y.dwdydydwThen, y 2so y 2. Then,dtdtdtdt7. y dy 1 ydtt12 dw y ydttdw 1 1 ydttdw 1 wdtty2ty2 t1 t1 .t R11dw w is µ(t) e 1/t dt eln t t, t 0.The integrating factor fordtttMultiplying through by the integrating factor and solving for w gives us:dw 11 w dtttdw w 1tdtd(tw) 1dttw t Cw Ct 1 1.Because w 1/y, y 1/w, so 1/y Ct 1 1 which means that y 1/(Ct 1 1)or y Ct/(1 Ct).11. Homogeneous of degree 013. Not homogeneous15. Not homogeneous17. The equation is homogeneous of degree 1. Let t vy. Then, dt vdy ydv.
20Substituting into the equation, separating and integrating yields2tdt (y 3t)dy 02vy(vdy ydv) (y 3vy)dy 02v(vdy ydv) (1 3v)dy 0(2v 2 3v 1)dy 2vydv1 2vdy 2dvy2v 3v 1 111dy 2 dvy2v 1 v 1ln y ln(2v 1) 2 ln(1 v) C2v 1y C(v 1)2(2t y)yy C(t y)22(t y) C.2t y y ty 2tAnother form of the solution is 2 ln ln ln (t) C.tt19. The equation is homogeneous of degree 2. Observe that either t vy or y ut results in an equivalent problem. We choose to use y ut dy udt tdu.Then,(ty y 2 )dt t(t 3y)dy 0(t2 u t2 u2 )dt t(t 3ut)(udt tdu) 0(u u2 )dt (1 3u)(udt tdu) 02u(1 2u)dt t(1 3u)du 02u(1 2u)dt t(3u 1)du13u 1dt dut2u(1 2u) 1 111dt dut2 u 2u 111ln t ln u ln(2u 1) C24 4 ln t 2 ln u ln(2u 1) C1 Cu2 (2u 1)t41y 2 (t 2y) Ct4t32ty (t 2y) C.
21 Another form of the solution is 1/4 ln t 2 yt 1/2 ln y t ln (t) C.21. y 3 (3 ln (t) C) t3 023. This is homogeneous of degree 1. (Also, observe that this is a first orderlinear equation in y.) Solving it as a homogeneous equation, we let y ut dy udt tdu. Then,(t y)dt tdy 0(t ut)dt t(udt tdu) 0(1 u)dt udt tdu 0dt tdu1dt dutln t u Cyln t Cty t(C ln t). 3y 2t t 2 y25. 2/3 ln 1/2 ln ln (t) Ctt27. The equation is homogeneous of degree 2. Let t vy dt vdy ydv.Then,y 2 dt (ty 4t2 )dyy 2 (vdy ydv) (vy 2 4v 2 y 2 )dyvdy ydv (v 4v 2 )dyydv 4v 2 dy14 dy 2 dvyv1 4 ln y Cvy4 ln y C.t y yAnother form of the solution is 1/4 ln ln (t) Ct ! 2 t2 t ty y(t 2 y) 3 1/3 3 arctan 1/3 ln (t) 29. y t, 1/2 lnt2tC 131. 1/2 ( y t) y 1 e t/y ln (y) C33. y 0 2y t2 y 1/2 is Bernoulli with n 1/2. Let w y 1 1/2 y 1/2 . Then,
22dw1dydydw y 1/2 2y 1/2. Substituting into the equation yieldsdt2dtdtdtdy 2ydtdw 2y2y 1/2dtdw y 1/2dtdw wdt t2 y 1/2 t2 y 1/21 2t21 t2 .2 Using undetermined coefficients, a general solution is w Ce t 21 t2 t 1.Thus, y (Ce t 21 t2 t 1)2 . Applying the initial condition results iny543y 41 (4 8 t 8 t2 4 t3 t4 ).21t 35. y 1/2 2 2 t2 t37. y 1/2 1/2 t2p339. y 3 ln (t) 061504100502-4-224t12345t41. y 4 dt (t4 ty 3 )dy 0 is homogeneous of degree 4. Let t vy dt
23vdy ydv. Then,y 4 dt (t4 ty 3 )dy 0y 4 (vdy ydv) (v 4 y 4 vy 4 )dy 0(vdy ydv) (v 4 v)dy 0v 4 dy ydv11dy 4 dvyv1ln y 3 C3vy33 ln y 3 C.t y 1 y3 ln ln t 8/3 ln 2 0.3 t3t 43. We need to solve dy/dt y/t t/y subject to y( e) e. The equation ishomogeneous of degree 2. To see so, we rewrite the equation:Applying the initial condition results indyy 2 t2 dtyt2yt dy (y t2 )dt.Now, we let y ut dy udt tdu. Substituting then gives usut2 (u dt t du) (u2 t2 t2 ) dtu(u dt t du) (u2 1) dttu du dt1dt u2 dut1ln t u3 C31 y3ln t C.3 t3 Now apply the initial condition and solvepfor y, y 2 t ln t. 45. Solve y 2 dx (x2 y 2 ) dy 0; y x2 x4 C47. The general solution is y 3 (3 cot x C) sin3 x or y 0. So, the solutionto the initial value problem is y 0.49. Because M (t, y)dt N (t, y)dy 0 is homogeneous, we can write the equationin the form dy/dt F (y, t). If t r cos θ and y r sin θ, dt cos θ dy r sin θdθ
24and dy sin θ dr r cos θ dθ. Substituting into the equation gives usdy/dt F (t, y) sin θ dr r cos θ dθr sin θ F F (tan θ)cos θ dr r sin θdθr cos θsin θ dr r cos θ dθ F (tan θ) cos θ dr F (tan θ)r sin θdθF (tan θ)r sin θdθ r cos θ dθ F (tan θ) cos θ dr sin θ drF (tan θ) sin θdθ cos θ1 dr.F (tan θ) cos θ sin θr53. f (t) t 1; g(t) t2 t; General solution: (tc y) 1 c2 c d 0 2d(ty 0 y 1) (y ) y 0 y ct 1 c c2 ; Singular solution:dtdtty 00 y 0 2y 0 y 00 y 00 (t 2y 0 1)y 00 0 y 0 21 (t 1) y 14 t2 12 t 34 .55. f (t) 1 2t; g(t) t 2 ; General solution: 1 2(tc y) c 2 y 1 2 2ct 1); Singular solution: y 12 (3t2/3 1)2 (c59. We see that the equation is a Lagrange equation by rewriting it in the formy ty 02 (3y 02 2y 03 ) and identifying f (y 0 ) y 02 and g(y) 3y 02 2y 03 .Differentiating with respect to t yields the equation y 0 y 02 2ty 0 y 00 6y 0 y 00 6y 02 y 00 and substituting p y 0 results indpdpdp 6p 6p2dtdtdt dpp p2 2xp 6p 6p2dxdx2xp 6p 6p2 dpp p22p6p p2dx 2x .dp p pp p2p p2 2tpThe solution of this linear equation is x y xp2 3p2 2p3 2p3 21p2 36p 6Cso6 (p2 2p 1)2p3 21p2 36p 6C 2p 3p2 2p3 .6 (p2 2p 1)61. Differentiating the equation gives us y 0 (2 y 0 ) xy 00 4y 0 y 00 . Now, we let
25p y 0 and solve for dx/dp:y 0 (2 y 0 ) xy 00 4y 0 y 00dpdpp 2 p x 4pdxdxdp2p 2 (x 4p)dxdxx 4p dp2(p 1)dx14p x .dp 2 2p2(p 1) This linear equation has solution x 83 34 p 43 p2 C 2 2p so p48 4y x(2 o) 2p2 1 p p2 C 2 2p (2 p) 2p2 1.3 33ky0k ay0a65. General solution:2t2 ln t t2 Ct2 . Initial value problem has two soluptions: y 2 t2 (ln 4 ln t).63. limt y(t) -464.65.yy44222-24t-42-2-2-2-4-4Exercises 2.61. 47.3742, 63.25723. 1.8857, 2.098475. 79.8458, 123.0487. 1.95109, 1.953889. 83.6491, 88.603511. 2.37754, 2.4189713. 185.34, 206.9814t
2615. 1.95547, 1.9560917. 90.6405, 90.692719. 216.582, 216.99223. 1.95629, 1.9562925-27. (a) y(t) e t , y(1) 1/e 0.367879h 0.1h 0.00525. (Euler’s)0.3486780.35848626. (Improved)0.3685410.36803927. (4th-Order RK) 0.4290690.41483129. y(0.5) 0.566144, 1.12971, 1.68832, 2.23992, 2.78297h 0.0250.3632320.3679180.36788Chapter 2 Review Exercises 325 t6 125Cdy113. The equation y y 2 is Bernoulli with n 2. Let w y 1 2 y 1 .dtttThen, dw/dt y 2 dy/dt so y 2 dw/dt dy/dt. With this substitution wehave,1. y 1/51dy 1 y y2dttt11 22 dw y y ydttt1dw 1 1 y (Divide by y 2 .)dtttdw 11 w .dtttRdw11An integrating factor for w , t 0, is µ(t) e 1/t dt eln t t.dtttdw 11Multiplying w by the integrating factor and solving for w gives usdtttdw 11 w dtttdw11t t w t ·dtttd(t w) 1dttw t CCw 1 t1 t C ytty . t C5. y 4 dy e5t dt 51 y 5 51 e5t C y 5 e5t C y (e5t C)1/5
27 2 t7. y (t) e e C9. For this first-order linear equation, the preferred method of solution is usingthe method of undetermined coefficients to find a particular solution of the nonhomogeneous equation. The corresponding homogeneous equation is y 0 3y 0which is the first order linear homogeneous equation with constant coefficients,y 0 ky 0, which has general solution y Ce kt with k 3 so y 0 3y 0 hasgeneral solution yh Ce 3t .Now, look at the forcing function, f (t) 10 sin t. Thethis forcing function is F {cos, sin t} and because neither ofof the corresponding homogeneous equation, we assume thatyp A cos t B sin t with derivative yp0 A sin t B cos t.the non homogeneous equation gives usassociated set forthese is a solutionyp takes the formSubstituting intoyp0 3yp (3A B cos t ( A 3B sin t 10 sin tso3A B 0 A 3B 10,which has solution A 1 and B 3. Therefore, yp cos t 3 sin t andy yh yp Ce 3t cos t 3 sin t.If you preferredto use the integrating factor approach, the integrating factorRis µ(t) e 3 dt e3t . Now multiply through by the integrating factor andintegrate the result. Observe that integrating the right hand side by hand involvesintegration by parts twice.y 0 3y 10 sin te3t y 0 3e3t y 10e3t sin td 3t e y 10e3t sin tdte3t y e3t cos t 3e3t sin t Cy cos t 3 sin t Ce 3t .11. The equation (y t) dt (t y) dy 0 is homogeneous of degree 1. Eithery ut or t vy result in an equivalent problem. We choose to use y ut
28dy u dt t du. Then,(y t) dt (t y) dy 0 u2 2u 1 dt t(u 1)du1u 1dt 2dutu 2u 11ln t ln u2 2u 1 C2 t 2 C u2 2u 1 t 2 Ct 2 y 2 2ty t2y 2 2ty t2 C. y t C1 2C 2 t2 113. The equation y 2 dt (ty t2 ) dy 0 is homogeneous of degree 2. We lett vy so dt v dy y dv. Then, substituting and separating variables gives usy 2 dt (ty t2 ) dy 0y 2 (v dy y dv) (vy · y (vy)2 ) dy 0(v dy y dv) (v v 2 ) dy 0(Divide by y 2 .)(2v v 2 ) dy y dv11dv dy y2v v 2 1 111 dv dy y2 v v 21 ln y (ln v ln(2 v)) C2 12 vln y ln C2vr2 vy C.vNow replace v with t/y and solve for C.r2 vvs2 t/yt/yy Cy Cy2 C2 t/yt/yty 2 C.t 2y
29 C C 2 Ct2Solving for y instead of C results in y .t x 2 t5x 2t1x5 ln (t) C ln ln15. ln 8t8t4t17. 13 t3 y cos t sin y C19. 12 t2 ln y y C221. y 1 Ce t /2 123. r t (cos t t sin t C)25. y t 1 (6et 6tet 3t2 et C)1/327. y 2 2 ln( 2/t), y Ct 2 ln Cp 3/21,6 t 6 t2 12 C29. y 1/3 t 6 t 6 t2 12 C 54p 3/2122y 1/3 t 6 t 6 t 12 C 54 6 t 6 t 12 C,p 3/21,y 1/3 t 6 t 6 t2 12 C 54 6 t 6 t2 12 Cp 3/21y 1/3 t 6 t 6 t2 12 C 54 6 t 6 t2 12 C31. y sin(t y) π33. t sin y y sin t 035. t ln y y ln t 037.
3039.1. V (x) Rx0πy 2 dt, W (x) ρV (x) ρRx0πy 2 dtF (x)A(x)where F (x) W (x) L and A(x) πy 2 . ρx3. y(x) K exp 2σ. If y(0) 1, then K 1.2. σ(x) 1 kx2 , x(0) 100, x(1) 60. x(t) kt Cso that C 1/100 and 31 1100k 1/150. When t 3, x(3) 150 100 3 33.33 grams.41.dxdtDifferential Equations at Work
31A. Modeling the Spread of a .40.40.20.20.00246810t0.0024B. Linear Population Model with Harvesting1((ay0 h)eat h) (d) limt aa 1 (e) limt h/a (f) limt ; y(t) 0 when t ln 1 y0ah2. (a) y 2 (b) y (y0 2)et/2 2 (c) (d) 2 (f) ; 2 ln 43. y 21 (2 et ) 0 when t ln 2 0.693; y 21 (4 3et/2 0 whent 2 ln(4/3) 0.5754. y 12 (2 et/2 0 when t 2 ln 2 1.3865. Firstsolve y 0 21 y 21 , y(0) 1/2 to obtain y 1 12 et/2 . Then, y(1) 1 1 2 e 0.176. Then, for year two, solve y 0 21 y 12 , y(1) 1 21 e to obtain y 1 2e(t 1)/2 21 et/2 so y(2) 1 2 e 12 e 0.938. y 0 11y 21 et/2 2r (2r 1)e(t 1)/2 so2 y r, y(1) 1 2 e has solution111y(2) 2 e 2r (2r 1) e 2 y(0) when r 4 ( e 1) 0.162. 6. Set r 0 in the above. Then, y(2) e 12 e 0.290. y(T ) e(t 1)/2 21 t/2 12 y(0) when T
dt dt 1 y 2 1 y 6 dy 8dt ln y 2 y 6 8t C y 2 y 6 Ce8t y 2 3Ce8t 1 Ce8t 1: 71. y exp c t (t 1) 73. L(t) L 1 e r Bt Exercises 2.3 1. The integrating factor is (t) e t. Multiplying through by the integrating factor, applying the theorem, integrating and solving for ygives us dy dt y 10 e t dy dt ye t 10e t d d
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