By Abell M.L., Braselton J.P.
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Additional info for Answers, .: to selected exercises for Introductory DE with BVP 3ed.
98 cv0 + 98 −ct/10 17. 6 so c = 5. 19. The parachutist’s mass is m = 192/32 = 6 slugs so we solve v = 32 − 12 v2 , v(0) = 60. Here, we use separation of variables: 1 32 − 12 v2 1 8 dv = dt 1 1 + 8+v 8−v = dt ln |8 + v| − ln |8 − v| = 8t + C 8+v = Ce8t . 8−v Now we find v: 8+v = Ce8t 8−v 8 + v = 8Ce8t − Cve8t v + Cve8t = 8Ce8t − 8 v= 8Ce8t − 8 1 + Ce8t and apply the initial condition v(0) = 60 ⇒ 8C − 8 17 = 60 ⇒ C = − C +1 13 17e8t + 13 . The limiting velocity is limt→∞ v(t) = 8. 17e8t − 13 dv dv dr dv 21.
Xi[n_]=x0+n h; yi[n_]:=yi[n]=h(f[xi[n-1],yi[n-1]]+ f[xi[n],h*f[xi[n-1],yi[n-1]]+ yi[n-1]])/2+yi[n-1]; yi=y0; xi:=n->x0+n*h: yi:=proc(n) option remember; yi(n-1)+h/2* (f(xi(n-1),yi(n-1))+f(xi(n-1), yi(n-1)+h*f(xi(n-1),yi(n-1)))) end: yi(0):=y0: The following implements the fourth-order Runge-Kutta method. xr:=n->x0+n*h: xr[n_]=x0+n h; yr:=proc(n) yr[n_]:=yr[n]=yr[n-1]+h/6(k1[n-1]+ local k1,k2,k3,k4; 2k2[n-1]+2k3[n-1]+k4[n-1]); option remember; yr=y0; k1:=f(xr(n-1),yr(n-1)); k1[n_]:=k1[n]=f[xr[n],yr[n]]; k2:=f(xr(n-1)+h/2, k2[n_]:=k2[n]=f[xr[n]+h/2,yr[n]+h k1[n]/2]; yr(n-1)+h*k1/2); k3[n_]:=k3[n]=f[xr[n]+h/2,yr[n]+h k2[n]/2]; k3:=f(xr(n-1)+h/2, k4[n_]:=k4[n]=f[xr[n+1],yr[n]+h k3[n]] yr(n-1)+h*k2/2); k4:=f(xr(n),yr(n-1)+h*k3); yr(n-1)+h/6*(k1+2*k2+2*k3+k4) end: yr(0):=y0: Answers, Hints, and Solutions to Selected Exercises 3, 11, and 19.
No. 19. First, we rewrite the equation as dy/dt − ry = −ay2 . Now, we let w = y1−2 = y−1 . Then, dw/dt = −y−2 dy/dt so −y2 dw/dt = dy/dt. Then, −y2 dw − ry = −ay2 dt dw + rw = a dt ert dw + rert w = aert dt (Divide by −y2 ) (Multiply by the integrating factor ert ) d rt e w = aert dt a rt e +C r ert w = w= a + Ce−rt r y= 1 a/r + Ce−rt (Integrate) (Solve for w) (Solve for y). Now apply the initial condition y(0) = y0 to obtain C = (a y0 + r)/(r y0 ). Thus, 1 r y0 y = a ay + r = . 0 a y + (a y0 + r)e−r t 0 + e−r t r r y0 21.