Give the correct answer and solve the given equation displaystyle{left({x}-{y}right)}{left.{d}{x}right.}+{left({3}{x}+{y}right)}{left.{d}{y}right.}={0},text{when} {x}={3},{y}=-{2}

Maiclubk 2020-11-24 Answered
Give the correct answer and solve the given equation (xy)dx+(3x+y)dy=0,when x=3,y=2
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Expert Answer

Sally Cresswell
Answered 2020-11-25 Author has 91 answers
The given diferential equation is
(xy)dx+(3x+y)dy=0,
with x=3,y=2 Now
(xy)dx+(3x+y)dy=0,dydx=yx3x=y
since the given differential equation is a homogenuous equation, putting y=vx, we have
dydx=v+xdvdx
So the equation becomes.
v+xdvdx=v1v+3
xdvdx=v1v+3v=(v+1)2v+3
v+3(v+1)2dv+dxx=0
v+1+2(v+1)2dv+dxx=0
dv(v+1)+2dv(v+1)2+dxx=0
dx(v+1)+2dv(v+1)2+dxx=C where C is a integrating constant
ln(v+1)=2(v+1)+ln(x)=C
ln(x(v+1))2(v+1)=C
putting v=yx we have
ln(y+x)2x(x+y)=C
Since x+3,y+2, we have C=ln565
Hence the solution of the equation is
ln(y+x)2x(x+y)=ln565
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