How Can You Solve for Limits and Discontinuities in Differential Equations?

[Mondon]

Homework Statement

\frac{dx}{dy}=\frac{k y-40\sqrt{x^2+y^2}}{k x}

Given the parameters conditions (0,0),(1000,0)

Homework Equations

substitution v=\frac{x}{y}

The Attempt at a Solution

\frac{dv}{\frac{1}{v}-v-\frac{40}{k v}\sqrt{v^2+1}}=\frac{dy}{y}

\frac{-1}{2}\ln({40\sqrt{v^2+1}+k+v^2k})=\ln{(y)}+c

k(\frac{x^2}{y^2}-1)+40\sqrt{1+\frac{x^2}{y^2}}=cy^{-2}

Sooo how could I possibly use my limits? I end up with a discontinuity at y=0 or is there some horrible mistake in my solution?

 

[Mondon]

That first Y should just be a y

 

[JDoolin]

It looks a little easier when I tried it. Sorry it's 5 months late!

You made the replacement, which you can do with "homogeneous differential equations" were the function of (x,y) can be expressed in terms of (v=y/x, usually) but you subbed (v=x/y).

v=x/y, which implies

y v = x, which implies

dx = v dy + y dv, and

dx/dy = v + y dv/dx

Now, we already have dx/dy, which is given as your first equation.

I replace x/y with v, and I got

dx/dy = v + y(dv/dy)=v-\frac{40\sqrt{1-(\frac{1}{v})^2}}{k}

and the v's cancel, so it might be a bit easier to solve.

 

[SammyS]

It looks a little easier when I tried it. Sorry it's 5 months late!

You made the replacement, which you can do with "homogeneous differential equations" were the function of (x,y) can be expressed in terms of (v=y/x, usually) but you subbed (v=x/y).

v=x/y, which implies

y v = x, which implies

dx = v dy + y dv, and

dx/dy = v + y dv/dx

Now, we already have dx/dy, which is given as your first equation.

I replace x/y with v, and I got

dx/dy = v + y(dv/dy)=v-\frac{40\sqrt{1-(\frac{1}{v})^2}}{k}

and the v's cancel, so it might be a bit easier to solve.

Substituting vy for x in the right hand side of OP's first equation gives: \displaystyle \frac{k y-40\sqrt{v^2y^2+y^2}}{k vy}

Then I get \displaystyle dx/dy = v + y(dv/dy)=\frac{1}{v}-\frac{40\sqrt{1-(\frac{1}{v})^2}}{k}

So that the v's don't cancel.

 

[JDoolin]

Oops, you're right. Probably much better to replace v = y/x instead, then.