Question

3. Given that yı(t) = t, y(t) = t, and yz(t) = are solutions to the homogeneous differential equation corresponding to ty" + t'y" – 2ty' + 2y = 2*, t > 0, use variation of parameters to find its general solution.

Answer 1

solution 23 y." + t²y" - 2 ty'+2y=2 tt, to Given that y = t = t2 and 2 = 1 are the solution to the homogeneous equation, then the complementary solution. is hy = c; y; + C2 Y₂ + (34z = C, t + C₂ t² + 2 / 1 By method of variation of parameter, the particular Solution is of the form p = Alay +B(X) Yz + c(2) Iz whese gt) Walt) Aco- grt) Walt) dt wit whese wilt ) = logi =-3 2t2 i git) W, (H) Wlt) I wit) 93110 t2 & Y = 0 et Ş" 2" 1 2 3 Yz 1 W2 (t) = 1 Vt y = 1 o -/+2 33" To 1 21+3 14 I ol It t² ol W3 (t) = y; I ol=1 at o=t² y yy ! O 2 Wit) = I y l It t2 t I = 1 et 142/ git)=2 tt

4 5:20:24 -- 0 = 1 * 2 d. 240 So, particular solution is Ip = - to xt + 2 ts y = - + + 2t to to I - T5 xt & 2 x 12o Then the general solution is y=&ctyp ly(t) = C, A + C₂ t² + + 7 12o