Differential Equation Help

  • Thread starter fiziksfun
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  • #1
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Homework Statement



i have the equation dy/dx = [tex]\frac{(y/x)-4}{1-(y/x)}[/tex]

i am told that v(x)=xy

and v=y/x

Express this using v, dv/dx, and x


I can get dy/dx = [tex]\frac{(v)-4}{1-(v)}[/tex] but that's it

Homework Equations



??

The Attempt at a Solution



I don't even know how to do dv/dx

can anyone help. im so confused?
 

Answers and Replies

  • #2
rock.freak667
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If y=Vx, then dy/dx = V+x(dV/dx)


[tex]V + x \frac{dV}{dx}= \frac{V-4}{1-V}[/tex]

make dV/dx the subject it now.
 
  • #3
quasar987
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Homework Statement



i have the equation dy/dx = [tex]\frac{(y/x)-4}{1-(y/x)}[/tex]

i am told that v(x)=xy

and v=y/x
Which of the two is it??

You are told that y is a function of x and that its derivative dy/dx satisfies

[tex]\frac{dy}{dx}=\frac{(y/x)-4}{1-(y/x)}[/tex]

But this differential equation is complicated. We would hope to simplify it by considering a new function v(x) that we would define as v(x)=y(x)/x so that the right hand side of the differential equation becomes

[tex]\frac{v-4}{1-v}[/tex]

like you said. So it is v=y/x that is useful for simplifying the equation, not v=yx.

To find dv/dx, use the chain rule. rock.freak667 gave you the answer but make sure you know how to obtain it yourself.
 
  • #4
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i don't understand how the chain rule applies in this situation!!

help!! how do i take the derivative of (v-4)/(1-v) with respect to x?? there is no x!!!
 
  • #5
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You are using implicit differentiation to get to what rock said, if you see how he got that, then just solve for dV/dx, at that point it's just a separable DE
 
  • #6
quasar987
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i don't understand how the chain rule applies in this situation!!

help!! how do i take the derivative of (v-4)/(1-v) with respect to x?? there is no x!!!
Sorry, it's not the chain rule, it's just the rule for differentiating a product of function. Anyway, you have expressed the right hand side of

[tex]
\frac{dy}{dx}=\frac{(y/x)-4}{1-(y/x)}
[/tex]

as a function of v, which you have defined as v=y/x. Now you'd like to express the left hand side as a function of v also. So you compute

[tex]\frac{dv}{dx}=\frac{d}{dx}\left(\frac{y}{x}\right)=\frac{1}{x}\frac{dy}{dx}-\frac{y}{x^2}[/tex]

Solving for dy/dx gives you what rockfreak wrote and now you've got a simpler differential equation in v as a function of x to solve:

[tex]
v + x \frac{dv}{dx}= \frac{v-4}{1-v}
[/tex]
 

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