Question

6. Solve the differential equation by variation of parameters. y" – 2y' + y = 1+x2

Answer 1

Variation of parameters method Consider second order differential equation with constant coefficients as: day tale + Qoy = Q(30) , Az dacz Az to Then the solution will be a y = Ye typ with complementary function Yc = G 9, +© Y. and parti cular integral Yp= 0, 4, +42 42 such that and ų =/ Wz de where is defined as a WL doc 9 1, =S W and Wz Wi 7 y, o Y Y Y and W₂= =ليا Wis Y, QUE Q(c) Ya Now, we have to solve the differential equation y" - 2y +y = variation of parameters y"-2y'+y= by HX3 14x2

day - 2 dy dax ex x dacz +y = H22 9 H22 for the complementary furetim Yc: Auxiliary equation of the differential equation will be m2_2m + = 0 (m-1)=0 m=1,1 So, Ye = (G+G2) e* or y = 9 taxe (+9 4, +542) 9 » Y,=é Y = xe² Now, determining value of w, w, and Wz as : ace² 42 ex و D) xe tex e e^(xe"te) - (xe 2x 27 2x 27 xe te - 21 /W = e 22 xex 1+x2 Next, Y2 Q(1) Y wi JQc0 е HX2 xe te

xe22 1 wi It x² 22 o Also Wat ex y, 0 y Q (0) ས་ ex Hx2 14x² -> W21 2x е H22 27 So, 41 = Will do af te ha dx da → 4 = 6 - 2 uke? 2 let 1+x²=t 2xdx=dt or xdx=dt dt at -/ logt So, 14 = -f [ 14 =- £ log (1H0²) 2x -da and the side of 14x2 eza → 4₂ = Stora de / dx = tants 14X2 b = tanta

Therefore, Yp = 4, 4 tylly, Y = et Y₂ = xet » Yp=e* (- log (143) + xe tanse Ye = Getxe or Yc = (a+Gx) et Hence, required solution of the given differen- tial- equation : y = Ye Up + y = G+qdJe + é 4-1 log (1+z2) + xet tanta or y=(G+qx - I log (1+x2)+x ton">)e* Please rate the answer.