Differential Equation Help

Homework Statement

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

i am told that v(x)=xy

and v=y/x

Express this using v, dv/dx, and x

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

??

The Attempt at a Solution

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

can anyone help. im so confused?

rock.freak667
Homework Helper
If y=Vx, then dy/dx = V+x(dV/dx)

$$V + x \frac{dV}{dx}= \frac{V-4}{1-V}$$

make dV/dx the subject it now.

quasar987
Homework Helper
Gold Member

Homework Statement

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

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

$$\frac{dy}{dx}=\frac{(y/x)-4}{1-(y/x)}$$

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

$$\frac{v-4}{1-v}$$

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.

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!!!

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

quasar987
Homework Helper
Gold Member
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

$$\frac{dy}{dx}=\frac{(y/x)-4}{1-(y/x)}$$

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

$$\frac{dv}{dx}=\frac{d}{dx}\left(\frac{y}{x}\right)=\frac{1}{x}\frac{dy}{dx}-\frac{y}{x^2}$$

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:

$$v + x \frac{dv}{dx}= \frac{v-4}{1-v}$$