Determining Linearity in First-Order Differential Equations

  • Thread starter Thread starter NJJ289
  • Start date Start date
  • Tags Tags
    Linearity
NJJ289
Messages
18
Reaction score
0
So I just started taking an intro diff eq course and here's one of my homework problems:

"Determine whether the given first-order diff eq is linear in the indicated dependent variable."

(y2-1)dx + xdy=0; in y; in x

I got the whole bit about the general form for linearity but I was thrown off by having just a 'dx' and 'dy' instead of the more familiar 'dy/dx'

(answer to the question is non linear when y is dependant, linear when x is dependant)

I'm confused as to what exactly 'dy' or 'dx' means, both conceptually and mathematically. I have a feeling there's a nice thread on this somewhere...

thanks for the help!
 
Physics news on Phys.org
Please note forum policy on homework.

https://www.physicsforums.com/showthread.php?t=88061

Having said that here is a hint.

\begin{array}{l}<br /> \left( {{y^2} - 1} \right)dx = - xdy \\ <br /> \left( {1 - {y^2}} \right)dx = xdy \\ <br /> \frac{{dy}}{{dx}} = \frac{{\left( {1 - {y^2}} \right)}}{x} \\ <br /> \frac{{dx}}{{dy}} = \frac{x}{{\left( {1 - {y^2}} \right)}} \\ <br /> \end{array}

Can you see which has x as the dependent variable and which has y?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top