Question

2. Solve the following Bernoulli's differential equations. (a) 2204 + 2xy – 2,2 = 0 (b) = 54 – 3y2 (c) y 2 x +48 = 1 (d) copy + 6y = 3x44/3 dr

Answer 1

* answer :- (a) given differntial ef is x2 dy +24a - 43=0 da ie uzdy tray= 43 - do which is Bernoullis differential eq x²dy +22y=43 dx Dividing both side by ya xzdy + 2 x 4 = 1 43 da B d22 ydy + 2 y = - 0 ut 52 = u biff w. dt. 2 -2 y 3 dy - du da da - y3dy - I dy i= ł da duzu de _ 4 u = - 2 2 da which is einear differential ear Here P(x) = -1 / f Qcx) = - 2 / 2

Spradda fa da Ize =e" = e 45 de - 4loge ze I = eloga - 7 = 4 > I- solution is u.I= (arn). I da to Cc is constant integration) u I = 6-2. I da to . a - = -2 nb dato -96+1 I = – 2 a - 76+1 to - 2 x 5 11 - to 0 tc exa + cx" y 2 = 2 x + (x" which given differ is required solution

(b) dy = 54 - 342 dy +34²=54 da dy - 5y=-34² da Dividing both side by y I dy - 54 =-3. 42 dx 42 :. 4²y'-syl = -3 put y'=u Diff-wort. 2, we get - da 52 dye-dy .- du – 54 =-3 - du +5u=3 da which is linear diff. eg Here P(x) = 5, Q(x)=3 I= Spranda S5dx 51de 52 = e = e = e : solution of L.DE is u. I = Sacx) I dato u. 23. 5x dx tc ex_3 esa - to

y e 3 + c 5 y = 3 otce sa IC dividing by both side by esa) which is required solution co) y ² dy + y = 1 42 dy do 1 -4 2 0 da Divicting both side by Yady + y = 1 - put 43=u Diss. wert a 34² dy _ du da da y dydd E90 become, I du tu=1 3 da du +3.4=3 Here It is L.DE P(x)=3, Q(x)=3 da Spaoda 32 エ = Soda e e = (I = 327

solution of L.DE is u. I=SQ12). I da to :. u. eBa_13. e3x dx + c 4 e 3 = 35e3x dx tc :. 43. 3x _ Xe to 43. 3 4 _ 3 to 1. y3 = 1+ c x | -- ( dividing both side by B) which is required solution (d) given Differential eq a dy +6y=32 43 de dy + y =3@y Edividing both side by x) > . (e dividing bothside by 3 dy + 6 to 4 = 3 your ydy + 4y43 With y dy+43- put y = u - 3 dydu ya d de

Eq' become. -3 du + e u=3 du - uz which is einear elifferential ear 10gx2 C = e Here pox) = -2 & Q60 = -1 I= S-2 do -25 62 -2logx 1= ht = I= x 2 1 2² 22 solution of LDE is u. I = SQ (and I dato CC j constant in 1. 1 2 = SC-D. (2) dx + c - Sazdxtc ...(:u= y + 3) 1 x -tc ) += + - y = x. x² + ca ² t=a+2which required sudion or
By using the definition of Bernoullis equation and using rules for solving Bernoullis equation.