Question

Solve the following equations using the indicated methods A] Undetermined Coefficients: y" – 2y + y = 621 B] Variation of Parameters: et y" - 2y + y = t

Answer 1

solution zt e y"-2y'+y = 0 Here given differential equation is y"- zy'+y 0 Corresponding homogeneous equahon Its characteristic equation is m2-2m +1 = 0 - (m-1)²=0 > m=lil are roots Thus solution of homogeneous equation is Yh(t) = (6) +(2+) et where G. (e are constante Now right hand side of 0 is ? Thus parbcular solution Yplt) = Ap²t * yocts Y»' (t) = GAB Use these values in 0 we get 4A2t_2 (246?t) + Ae?t at of 0 is zt ZA 626 at e A et zt A=1 Thus Yplt) = ezt Thus nowo general solu bon of a becomes Yhlt) + Yplt) + [yet) = (CI+(at) et + e? where C, (z are arbihury constants. YCH) -

به اند t i solution Given differential equation is, t y - 2y' +y = e Auxiliary equation of homogeneous differential equation is m?- 2mti = 0. cm-1)2=0 ism1,1 of homogeneous equation is, Yn (t) - Cat Gt) et where c, and C₂ are arbitrary constants "Y, (t)= et and Y, (t)- tet are independent solutions of homogeneous differ- ential equation... Yp (t)= u(t):9, +42(t) yz(t). "Wronskian of Y, and Ya is, WCE) = (HE) e²t_tezt. et linearly et tet CHE) et is, :: WCt) = ezt. ... Particular solution Yp (t) = 4Ct). Y, Ct) + 42Ct). 42(t). and (t) = Yz(t), et tet. et teet -t7 dt. WCE) luct) = -t щ с

Malt) - 5 4. Ct). de set. et dt WCE) t ezt Particular .: 42(t) = 5 + dt log (t) solution Yplt): -tet + t log Ct).et General solution of given differential equation is, 9 (t = 9,(t) + Уr (2) catat) et = tett tet log (t) (citat) et+ (Jog(t) -1) tet Tyc+).