Question

Solve the following differential equations

x²d²y/dx² − xdy/dx − 3y = lnx/ x , x > 0. show that the answer is y = A/x + Bx³ − lnx/ 6x (2lnx + 1)

x d²y/dx² − dy/dx + 2y/x = (ln x)² .show that the answer is y = x { A cos(ln x) + B sin(ln x) + (lnx) ² − 2 }

Answer 1

Solutions 2² dy -ndy -3y= lan noo da? n 27 day lnu ,270 da? 2 This is a day 3y= a evler Cauchy equation which is typically solved by change of variable n=et ne-tal 3) Aza Then we have dit net die eerste rondezt substitute into the initial D E. , we get 20 ( - , -alby Jet-34.ptet dzy de 2 - 2 dy - 3y = te-t y" - 24"-3y=tet Complementary function m2-2m-3=0 m2-3m +M-3=0 mlm-3) + (m-3)=0 (MH) (m-3)=0 TM=-17, 1m33] Se, complemelony function Ye= Yc = Aet +Best are- Ae-t +Best

for particular integral Y,=e-t x stet Izre3t Walet est = Yezt neot 3e3t Now 92/4, X Yp =-9. y ₂ x dt W et tet dt 4 ezt ter de € + festetten Yezt ***2 + queferte de (-14++))e=441 36 Yp te 16 Yp tt? 8 -tet e-t 16 64 t é $64 16 y = Ye + Up y= A e-t + Belt +tet tet 8 =(A-14 Je + + Be cet Best ttent 2 yo 3t + t ct e tet 16 Yo = teet 8 16 pot (x=e] & [lana Bx² + ( lnx)² lux your t 82 162

solution =(nn)? ad y como o The 22 dy ndy +2y= da2 allax)² due du This is a ever couchy equation which is typically solved by change of variable t » xe-ti o Then we have . & dy da day de 2² dy e to do dy Jezt substitute into the initial De we get *(*1+ yo et (41" d24 dd dt² dt y" 2y + 2y = Complementary t^ on solving m=lt Ye = et [A last +Blent] Msaj corpámentary fenchonko ye -et (Alast +Blint) + 2y = e(t )2 et les m² -- 2m + 220

for particular integgel Yp = (At+84 +chet 4$=(2At + 8) et + (1+2+8++Cjet Yp - (za) et + (zat 48) et + (pat + B )et + (At?487+00et (ZA JE* +2(28/6)'ét+(at 4644 chet - 2004+Bet-2044"ftillet + 21487/877chet =ét (42) (At?) et +LA+K) tet + cet-ett? solving et² on ! A=1 tet ! ZA+6=0 = -2 So, Up= (+²20) et So, y = Ye + Yp [A cast & Bfint + 2²-26 put Tetra y=et & lt-lna y = ag n coolera) + Bilen (enn+(@mug?-2 Resea ? y= n { a cos clana)+ B Sencha) + (dua)?-23