Question

Need 20 and 22. please explain.

In Problems 19–26 solve the given differential equation. 19. (y2 + 1) dx = y sec? x dy 20. y(In x – In y) dx = (x In x - x In y - y) dy dy 21. (6r: + 1)y? + 3x² + 2y3 = 0 dx dx 22. dy 4y2 + 6xy 3y² + 2x

Answer 1

ор we ave , Яди – = xp (һих Жp (f-hик-хит к ) (1) Саи be written as dy чеим - — (2) с h noh - ћих и ау. и их- Put y= ux 2) у vt a du dn ax put in (2) °. E utn av о Кии – vueucru) и их – х «и(v1) {vu) ytoy- venen)- veneva) eniu)-encua)- venou)- veniva) - у encul-eneral-v лау = и nav = venga) -venttin) -venstvencontra an aniu)- lvlun)-v

nav enowl- encuad-v Now eniva encv) tencul U2 n av OLX PUIÇK) – ener) -en-V n av an 42 lolu (Vtv) encu) + V) dva -1 are V 2 teniv) t OVE - 1 an v2 $ Integrating both Siales, we have -Incul + enu -un in) + с V V Put dniu) =t - (*) ve et Differentiate (*), we have Idvedt Now, tutu) dva Edv I dniu) 4 2 V It V dt = ftet de ote-t-e-t --nov) e aniv) e-enco)

-enew) Э -env — Т. tin v = ди. + С Put V = y ture have и у -Ди (ч) - 4 и + ul 4 ) =enalto 4. — И Luly) + yent Ч. ) ylenin)+c) = || -у ди ич enculte - en = ot Нуги) = - и Хи (+ (17) forfatur-talno t solution where asbitary constant

PLEASE GIVE THIS A RATING IF THIS WAS HELPFUL FOR YOU.