Question

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10. y" – 2y' + 2y = 2x, y(0) = 4, y'0) = 8.

Find a solution

Answer 1

- equatich is, Here given given differential equation is yol-25'+ 2y = 2x, y2O) = 4, so, Homogeneouas equation is, yl-2y+2y=0 =) Su, Auscillary &²_20+2=0 o comparing above equation with cao2+ brt c=0 we get, a=1, b =-2, C=2 so, By formula method for yools of quadratic equation, we get, b²_uae -(-2) + (-2)2 - 2(1)(2) so, bt 11 2a 201) =) at ✓ 4-8 2tr-u 2 atdzi Elti 2 =) g=iti = 2+BI =) 82=1, 1 = 1 2 So, QY ICO) =) se, roots of companings roots et auxillary equation aire complexe so, complimenteery solution is given by, ci cos ( 13x) + C2 e sin (132) Yc (3C) = ciecos 20 + cz et sin (v 6:4== 1) Now, for perticular solution, we will use Method of usooder abel undetermined coefficients since Rostad RH.S. = fracy so, Perticular solution will be on the form Here = 2x

/ D doc =k do doc - for any constant k Yp(x) = AX+ 13 where A and 13 are constants. Then, yb (c) = dia K(1)-K and yo'cx) cu ( dk =0 for any consternt k 1) dic so putting values of yll (1) y (1) and Upeses in given differential equation, we get, O-RA + 2CA3C + 3) = 220 E 2A +213) + 2A) = 2X =) -2A +213= 0 and 2A=2 A =) so, 213 = 2A = 2013= 2 =) 3=1 ypca) xt is pesticular solution. so, solution or given CDE is, =) Yol Yola) y (x) = Yc (X) + Yp (30) Усх) cietcos (0)+cze since) + 30+ =) Now, yco)=4 gires, YCO) ce (US (0) + (z eºsin (es total - 4 CCI (1) + C2 (c) + 4-01= -) = 3 so, y(x) ze cos Ox) +C2 e sin(x3 + x+21 = Now, y'(x) = 3[ex (-sin(x)) + a em coscx)] + C2 [ e COSCX) +exsin (~)] +01+o

(: ਹੈ Cuca).VCX)) uer) et oss + vek).cluor) de aer, d sinox) = cusci), alcosce sincx) and ar dx =CD) doc and dk=0 for any constant k cx so, y'(x)= ze" cusco) Cus(M) - 3esincx) +€ Cz [ersinec)+ e* cos(x)] +) Mow, ylco)=8 ylco) = 32° cus (C) - 3eºginco) + C2 Teºsincat e casco)] +1 -) こ) ylco) 3@(1) (1) – 3 C1 (0) + C2 [(1) C6) + (101) ti =) 8 = اے =) Car 3-0 +2(1) + 8=40+C2 =) (2=8-4=4 555 so, solution of given Initial value problem is, Bex [ 3 cos (c) +4 sincx)] +x+i yox)