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melmel
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what do you get if you re arrange to dy/dx=
sorry but i dont know .................it equal to 1 dont know if I am correct >.<
after your get the left hand side to be dy/dx sub v=y/x
\[\frac{ xdy }{ y-xdx }=1\]
yeah, now move the other terms to the other side
\[\frac{ dy }{ dx }= \frac{ y-x }{ x }\]
\[(y-x)/x = y/x - x/x\]
x/x simplifies to ...
good so you have \[\frac{dy}{dx}=\frac yx-1\] now let \(y/x=v\) \(y=vx\) \(y'=?\)
use the product rule to find y'
\[y'=v(1) -x\frac{dv }{ dx }\]
should be plus not minus
now sub this in \[\frac{dy}{dx}=y/x-1\] becomes \[v+x\frac{dv}{dx}=v-1\]
which is now variables separable
\[\frac{ dv }{ dx }= \frac{ v ^{2} -v}{ x }\]
um you should have taken away the v from both sides, before dividing by x
$$(x-y)\,dx+x\,dy=0$$note this is homogeneous; both our functions \(x-y,x\) are homogeneous. as @UnkleRhaukus stated, use a substitution \(y=vx\) i.e. \(dy=\left(x\dfrac{dv}{dx}+v\right)\,dx\):$$(x-vx)dx+x(x\,dv+v\,dx)=0\\(x-vx)\,dx+x^2dv+vx\,dx=0\\x\,dx+x^2\,dv=0\\x^2\,dv=-x\,dx\\dv=-\frac1x\,dx\\v=-\log x+C\\y/x=-\log x+C\\y=-x\log x+Cx$$
http://en.wikipedia.org/wiki/Homogeneous_differential_equation#Homogeneous_functions
guys can be this one \[xdv=(v ^{2}-v) dx .....then....... (v ^{2}-v)-xdv=0\]
then identify if its exact DE or not
\[v+x\frac{dv}{dx}=v-1\\ x\frac{dv}{dx}=-1\\dv=-xdx\\ \\∫dv=-∫dx/x\]
What is the second thing? where does the dx go?
v=- lnx +c is that right?
yes, now remember v=y/x
then sub v= y/x to v =-ln x +c
thats better \(\checkmark\)
tnx i other one here tech me ?
is there an equals sign somewhere?
\[\frac{ dy }{ dx }=\frac{ -x }{ y- }\]
ok this one is the same technique, rearrange to dy/dx= and substitute v=y/x
divide both sides of the fraction by x, to get it in the right form \[\frac{-x}{y-2x}=\frac{-x/x}{y/x-2x/x }\]
always? v=y/x even to other problem?
When we can express the DE as y'(x) = f (y/x) we can call this a homogenous equation and v = y/x substitution will work if we can't get this form y'(x) = f (y/x) this technique wont work
can you gve to me the final answer here this will serve as my basis
sometimes a 'homogenous equations" means something else
you gve a sense to me :)
i haven't worked out the final answer yet
by the way thank you :))) tnz alot