Question

Find y(t) solution of the initial value problem 2 y2 +bt2 y'= y(1) = 1, t>0, ty

Answer 1

112) Hene y = 292 1 bt2 ; Y(1) =Z,, t>0 ty i dy dy = 2외자 642 dt ty 292 bt2 ty : - 2 + b Let Fly, t) LI 32 IT 12(월) 2(53)746 () : Fis Homogeneous fumction of degmee zeo. To find solution we tuke VE 9. t ' y = vt dy 2 t, dv. dt dt

Substitue all values wc get, V 11 t. do 11 vtt du 202 + b dt V t du 222 + b dt t. do = 2 v2 + b-v2 dt v2 tb dt V dv = at 22b d 7 Integruting both sides we get, 2 l 20 do = dt 22+b t. 1 log (v2+b) = log t + loge log tt logc , c = constant 2 abitinn Cabocah I log (w2+b) = logit.c) ley.cv27 b) 2 log (4.c) log (w2+ b) = log (t.c)² 2 v2+ b = Ł2 (2

.) ² + b = 2² 2² y² +bt2 = +42 - (1) we get Substitute t = 1, y = 1 in ean (1) 11) ²+ b(2)2 = (2)4 (2 1 + b = c2 Substitute c2= 1+b in ean (1), we get y²+bt² +4 (1+b) y2=th + b +4-bt², which is required solution of given TVP.