Question

Question completion Statue Problern 2a. (7 pts) Solve the following DE. V" + 8y" 62 +9+2, Problem 2b. (8 pts) Solve the following DE: y' +2y +y= e In(a) Problem 2c. (10 pts) Consider the differential equation rạy" – (+ 2x)y' +(+2)y = 0. Verify that y1 = r is one solution of this homogeneous differential equati find the general solution of the following DE: x?Y" – (.r+ 2x)y + (x + 2)y = 23.

i need all the question with steps .

2)a) last number is 2 ,,,,it is blurry in the photo

Answer 1

8. 3 a) +8y" = -622 +9X +2 y" dy dx3 +8 y dx² ~6일 4342 (' + 803) } 6일+92 where d dx AE M3 t 8m2 = m2(mt8) =0 m=0, -8 0X + Xeox + Cge3x qt N2 + ge-8x (-62-9242) 03커 8D2 PT. 1 85 (14g) (-404 2143) 5. [1+ ] [-Cx2 + 9x+2] 1 8D2 8D2 [ 1- 올 + ] (-12 + 2x+2) [-622 49Xtg 능(-127 +9) + 5 [ -z + + ] 1 802 (디 (12)] 증 ' + g + + X 22 2 12. . 16 2 + 섎 22. 32 256 크 23 + 10 solution: CF + P.I. Gta t se8 t 기자름기 중에 11 25G

(2) Ы 8" + 2 + e-x lnca) 2D4+1) é x ln(x) d where D = А.Е. m2 +2m+1 а (m+) =0 СЕ. Gext Guex To find P. I. we use variation m= -0, -1 Parameter methal het e-1, Я, = e-x * = e- Х Хе-х |-- - е-2ҳ +о e- — е-х tex NOW Р. І. : 0 4 + 4, 9, НІ where us 8. ( x) da W u = xen en ln(x) dx e-37 U | a ln(x) da т а, е - Зоя) - | ; ; | 4 94(7) + 11 2 2 - - 9n () + - + ч 9 t 4 NOW І! W

с e- u, e-n ln(x) | | da e-2 Ug = (1. ln (X) de д т О. 1. Ау " , ln (X). X Г U₂ aba) Hence І. 2 2 - 9. а) ++) е* + ( 13 (1) -х). зех - 1* ex 96 (7) + + е* + + * + *** 2, tx) -хе-х 1 x-ex 24 (1) 4 - с. г. х*е-х Solution та + Р. І. 4 = се- + ( хе-х t 12 е-х 9%Cx) - 3x-ex. А 2C) Ч = a the Chien " - (1437) + (+25% = о solution в the Ф.Е. Then 와 must satisfies Equation. я? " (? 4 2 ) + 4+ (х+2) 4, * xo (я° 427) x1 + (х + 2) хх - - 2 + * + 3 о Hence h= и the solution of DE

Now Second solution: find Given x²y" (x²+2x)) + (x+2) Y 73 Compare P(x) y" + xy + xy. 262) = (x² + 2x) Hx) Pla 90%) = (+2) 2 HR) Now find sext [-s(ebay roxy) dz] de (4) ² (2) da = P(x) s epicos du = 5 - x² + 20 de - Se + 2) de - [ x + 2logx] In de] exp [at Blogx] exp[ nt log (x2) exp [-S = Pla) ex. elog(x²) a²ex x²ex da V- Sexdx = ex Hence required Second Solution v.y = xen So C.F. antener Now to find use Variation Parameter. Lel y = x Y = nex ५, ५०

xen •ея xex tex Р. І. и, 9 + a, Ч, 4. а 4. (1) da N и, а 3 xe”. * ex da а, 2 -|- аж. - 3 3 2 9. - 20) 4 u, 13 ес E 2 |1'e-* 4ч. и, . -1°e-1 - Газе-1 4 - х*е-1 + 2e1 - 2 | e dy — и е-м+ 2 е-х + 2e-X ча Р. ія и, 4 + 4, - 1 + (-1°e + 2хе-х 4 2e-1) x xe+х - + (- 3 + 2х + 2x) 에초 + э 2° +2х. - ч 3 solution, = С. Р. + P. I. }= (х + схе* — - x2 + 2* + 2x .