Differential Equation Help

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.

 

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.

 

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

 

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]