How Do You Solve a Differential Equation Using Substitution Methods?

[fiziksfun]

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

Homework Equations

??

The Attempt at a Solution

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

can anyone help. I am so confused?

 

[rock.freak667]

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 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.

 

[fiziksfun]

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!

 

[NoMoreExams]

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]

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

<br /> \frac{dy}{dx}=\frac{(y/x)-4}{1-(y/x)}<br />

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:

<br /> v + x \frac{dv}{dx}= \frac{v-4}{1-v}<br />