Question

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7(0) = -5. Consider the initial value problem 47" + 28y' +49-0, (0) - 1, (a) Solve the initial value problem. X(t) Plot its solution for osts 5. (A computer algebra system is recommended.) (b) Determine where the solution has the value zero. (c) Determine the coordinates (to, Yo) of the minimum point. (to yo)-( (d) Change the second initial condition to y(0) -b and find the solution as a function of b. Xt) Find the critical value of b that separates solutions that always remain positive from those that eventually become negative

Answer 1

Solection:- 4y" +284' +49470 iyo = 4.34'10:51 Lii). Aurilory equation of equation (i) is 40m2 +28m +4930 if Cx2+bx+c co therr -X=-6F 62-400 > 2a ms- -_-28 + $(98)2-464)(49) 264) = -28 +17844784 m=28+o m==ž Hence General solution is included only. complementary you function. Tyyat general :.4H) = (A+B+)e It is general solution. '(0)=) , Ped it in coor City weget 2-kt Boje 3.0;

Differentate equation (W) with respect to of we get glu) = Be it + tet) e 3+(-3) 9'45 **(0-3(n+et)] Put your ss inabore en weget -5= 2 3toj[B-1(1+2.0)] 1. A=1) -5= B-Z ::8= -5+1 = lep? -- B = -24 ] peet Aal eB= 3 in equation (i), weget a y(t) =(1-3 +3 € 3t Required solection of emotion For actes 04+: o 0:24 Tit -027 -ot

(6) ро» 800 «о = (1-3te to -(4-34)=0 - 3 et 1 Tit= } ] Hence at Solection yt has the zoro ralue