Question

Solve the given differential equation by variation of parameters. 2x²y" + 3xy' - y xi

Answer 1

OS - 2x2y"+3x4'-y = x35x solution to homogeneas port 21²4" tany'-y=0 (40) = untczaka particular solution using Varaten of parameter et Yp=4.J1 + 4292 where. I1=%, 42 = x2 192919) -Swahid -cauchy equatia =) Z=lna Os UL da U2 = ida x=eZ xy'=dy dzi dy dy de W(7) (1)M YI krnl= Y ZP yil Y 312* 312 # 2024 xt xiz '+3241-y=0 atel in dy_y=0 -222 =) 2¢y zdy +3 dz dz2 Iz W(x) = 2 2222 て adly Now, us= ak. 3 X32 da + dy - dz2 dz Charectenstice equatia 2r2tr-1=0 272+27-Y-130 2r1Y+1)-1(r+1)-0 (28-1) (r+1)=0 rt120=1:-) 28-lat *2*21*02 = -7/ 3 / 2 4 2 2x x 24 -2 x %+ nin こ -4 x min so 39 292?) zu 25=zh zz -Z Azz art aha x= 2pt. z

Y1 g(x) du W(0) * x3ja da og 2 x 3/2 s x32.x²Jse da = 2 // Jarda 2 3 PS = ? as IS How, p=4291+4242 = 7 2 2 + 2 x 5 x 2 39 کر) & ak --- -- 39 + 2 x - -20 +262 195 6 all/ 195 19/ - 2 65 2 9р so, the gensed solute y = Yet Yp 19: (2x* tq ka + 2 h