Question

Solve the given initial value problem. = 4x + y - e (0) = 2 dt dy dt =x+uya (0) = -3 The solution is x(t)= and y(t)=

Answer 1

solution :- Given = 4x+y - 4t da at dy dt = 22+ 4y ; 4(0)-2,400)=-3 This can be written in Matrix form as e46 2 X'= 54 where [] X de la and х dt dy at we to 2 First solve the homogeneous equation corresponding which is x'a Х 4-7 - 1 [* 4] Eigen values for [* d) is, the roots of /4-2 | ie, (4-2)²-1=0 16-82+2-120 ie, 2²-87+15=0 (2-5) (2-3)=0 2=5,3 Corresponding to 7=5, lo ie, (3) 1 1. 4-7 (4.2 2003 implies (-)[:)-(3) -- de, 2-5 So x+y=0) R=4. eigen vector corres. corresponding [:] an 13

[i][b]-[: il aty zo vector corresponding is an [ . to vecters t ces Czest y=c, este When a=30 ③ becomes, il, x=-4 So eigen a=3 For distinct eigen values n., ..., an and eigen ni...) "n the general solution takes the form en tu + C ₂4 M + + C, е man x =ç e Here 8,25, ^,- () m, x=c, e solution of homogeneo to azt 22=3, 2 So [:] general homogeneous equation corresponding 54 4 2- e 3t is the can we a Now use method of undetermined Coefficient for finding particular solution of @

we let X is a of particular solution (Here C, and Ca are constants to be determined.) is the non past. Here [osta] shomogeneous From 6, X = get *CE Putting ② 46 in ③ 14€ 4e4t *NE > +46 e- e 4t 4e a 4t 4 e C (4G+ (2) eft 6, + 4cm) 84 70 49 =4C,+C2-1 ***** 4C2 = 4, +462 te) q=1 c=0 ;. From 5 **03-627) is a particular solution of Hence general solution of ② is, zest By adding Х = /cest and 6 3t ut çe 5t + C₂ e te

36 ses 8 2 Thus XC+) = ciest y(t)= C, est+cest text is the general solution of Now to solve the given IUP, we use the given conditions, 2 (0) = 2, y(0)=-3. From a(o)=2 a=0,-& C-C2=2 y (0)=-3 -3-C,+C2+1 = C,+C =-4 .:20,22 ie, C,=-1/ :, C2=0,-2=-1-23-3 3t -e + 3e st +3 e te g(t) = -25% is the solution of given IUP. 4+