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melmel Group TitleBest ResponseYou've already chosen the best response.0
bernoulli's Equation
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
what do you get if you re arrange to dy/dx=
 one year ago

melmel Group TitleBest ResponseYou've already chosen the best response.0
sorry but i dont know .................it equal to 1 dont know if I am correct >.<
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
after your get the left hand side to be dy/dx sub v=y/x
 one year ago

melmel Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{ xdy }{ yxdx }=1\]
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
yeah, now move the other terms to the other side
 one year ago

melmel Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{ dy }{ dx }= \frac{ yx }{ x }\]
 one year ago

melmel Group TitleBest ResponseYou've already chosen the best response.0
Im correct?
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
\[(yx)/x = y/x  x/x\]
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
x/x simplifies to ...
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
good so you have \[\frac{dy}{dx}=\frac yx1\] now let \(y/x=v\) \(y=vx\) \(y'=?\)
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
use the product rule to find y'
 one year ago

melmel Group TitleBest ResponseYou've already chosen the best response.0
\[y'=v(1) x\frac{dv }{ dx }\]
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
should be plus not minus
 one year ago

melmel Group TitleBest ResponseYou've already chosen the best response.0
ay sorry about that
 one year ago

melmel Group TitleBest ResponseYou've already chosen the best response.0
y′=v(1)+xdv/dx
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
now sub this in \[\frac{dy}{dx}=y/x1\] becomes \[v+x\frac{dv}{dx}=v1\]
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
which is now variables separable
 one year ago

melmel Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{ dv }{ dx }= \frac{ v ^{2} v}{ x }\]
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
um you should have taken away the v from both sides, before dividing by x
 one year ago

oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.1
$$(xy)\,dx+x\,dy=0$$note this is homogeneous; both our functions \(xy,x\) are homogeneous. as @UnkleRhaukus stated, use a substitution \(y=vx\) i.e. \(dy=\left(x\dfrac{dv}{dx}+v\right)\,dx\):$$(xvx)dx+x(x\,dv+v\,dx)=0\\(xvx)\,dx+x^2dv+vx\,dx=0\\x\,dx+x^2\,dv=0\\x^2\,dv=x\,dx\\dv=\frac1x\,dx\\v=\log x+C\\y/x=\log x+C\\y=x\log x+Cx$$
 one year ago

oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.1
http://en.wikipedia.org/wiki/Homogeneous_differential_equation#Homogeneous_functions
 one year ago

melmel Group TitleBest ResponseYou've already chosen the best response.0
guys can be this one \[xdv=(v ^{2}v) dx .....then....... (v ^{2}v)xdv=0\]
 one year ago

melmel Group TitleBest ResponseYou've already chosen the best response.0
then identify if its exact DE or not
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
\[v+x\frac{dv}{dx}=v1\\ x\frac{dv}{dx}=1\\dv=xdx\\ \\∫dv=∫dx/x\]
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
What is the second thing? where does the dx go?
 one year ago

melmel Group TitleBest ResponseYou've already chosen the best response.0
v= lnx +c is that right?
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
yes, now remember v=y/x
 one year ago

melmel Group TitleBest ResponseYou've already chosen the best response.0
then sub v= y/x to v =ln x +c
 one year ago

melmel Group TitleBest ResponseYou've already chosen the best response.0
y = xlnx+c ?
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
almost,
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
the last term,
 one year ago

melmel Group TitleBest ResponseYou've already chosen the best response.0
sorry about that
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
thats better \(\checkmark\)
 one year ago

melmel Group TitleBest ResponseYou've already chosen the best response.0
tnx i other one here tech me ?
 one year ago

melmel Group TitleBest ResponseYou've already chosen the best response.0
xdx + (y2x )dy
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
is there an equals sign somewhere?
 one year ago

melmel Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{ dy }{ dx }=\frac{ x }{ y }\]
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
ok this one is the same technique, rearrange to dy/dx= and substitute v=y/x
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
divide both sides of the fraction by x, to get it in the right form \[\frac{x}{y2x}=\frac{x/x}{y/x2x/x }\]
 one year ago

melmel Group TitleBest ResponseYou've already chosen the best response.0
always? v=y/x even to other problem?
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
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
 one year ago

melmel Group TitleBest ResponseYou've already chosen the best response.0
can you gve to me the final answer here this will serve as my basis
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
sometimes a 'homogenous equations" means something else
 one year ago

melmel Group TitleBest ResponseYou've already chosen the best response.0
you gve a sense to me :)
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
i haven't worked out the final answer yet
 one year ago

melmel Group TitleBest ResponseYou've already chosen the best response.0
by the way thank you :))) tnz alot
 one year ago
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