IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

v_gurucharan

solve--------

(3xy+y²) dx + (x²+xy)dy=0

sol: x^2(y^2+2xy)-C=0

neeraj_agarwal_1990

its a homogenous DE....i suppose....

dy/dx = (-y/x) . [(3x+y)/ (x+y)]

now divide numerator and denominator by x in rhs...

and then put y=tx....

raulrag009

find dy/dx and put y=vx

substitute

dy/dx= v + xdv/dx

u get

{(1+v)dv}/v²+2v = -2dx/x

integrate both the sides

1/2 [{2v+2}/v² + 2v ] dv = -2lnx +lnC

1/2ln(v² + 2v) = -2ln(x) + lnC

thus

(v² + 2v)^1/2 = Cx^-2

^{substitue v=y/x}

(y²/x² + 2y/x ^{) 1/2}=C/x²

from this u get ur answer!!

v_gurucharan

show all the steps from the beginning!!!!
how did u get

{(1+v)dv}/v2+2v = -2dx/x ??????????

punnima



its a homogenous DE.. dy/dx = (-x²) ( 3{y/x)+{y²/x²}) --->(1) (pulling out a x² and x² common (x² ) (1+(y/x)) each from num and denom)

put y=vx

which on diff w.r.t x gives

dy/dx= v + xdv/dx

subs. d above in 1 u get

v + xdv/dx = 1. (3v+v²)

(1+v)

u get

xdv = -(3v+v²) -v(1+v)

dx 1+v

= -3v -v² -v-v²

1+v

= -2v(2+v)/(1+v)

dv(1+v) = -2dx

(2v+v²) x

integrate both the sides

1 ln(2v+v²) = -2lnx +lnc1

2

ln (2v+v²)^1/2 +ln x²=lnc1

(2v+v²)^1/2 .x² =C (raisnig both sides to e)

from this u get ur reqd soln

x^2(y^2+2xy)-C=0

punnima

hope my method is crrct........if not plizz tell me wher ive gone rong