Question

9. Write the solution to 2x²y" +22xy't by=0 for L= 42, 50, 58?

Answer 1

m-2 Given 2x 7" +22xy +YL = o. L342,50,58 ? het dessume x - xm Y = mam; y = m(M-12 Substitue of, y'y" in Abore equation i 2x² m (m-1) M-2 +22x. ment, am. L=0 2 (mumam + 22 m n = 0. xm [ 25 - 2m + 22m +2] =0. [ 2m +20 +1] -0 Roota of 2m? + 20mth so is for L=42 , 2 t.20mtha=0 m.- - i m2-- for different scoots me, ma a is mix man. Solution -32 72 + Ge But am=y Y ce te th th L=42 I final solution -3x

-52 for L=50 2m²+ 20m +50=0 Roots m, = m₂ = -5 for dildo Game Roots m. = ma solution is (arxea) me = [, +25] e final station Y = [C+ca z Jese for 2-50 L=58 2m²+zom +58=0 m = -5+2i ma = -5-2i ie m= 5=2i for Imaginary Roots &, tip solution is ea [c, cospa + sinpe) so 2=-5 B=2 am = 50 [c, as 2x+c, sin 2x] for Y = e 52 [c, cos2x+c sinza] final solution for L = 58.