Is the Differential Equation (y^2 - 1)*dx/dy + x = 0 Linear?

[RiceKernel]

Hi, I was just wondering if :

(y^2 - 1)*dx/dy + x = 0

In this case, x is the dependent variable.

Is linear? I know it is but I want to understand why. My question is with the coefficients.
The first coefficient has a y to the power of 2 to it and a constant. It is also a function of y (independent var in this case) for obvious reasons. I'm wondering if the power in the coefficient matters ? For example , instead of getting (y^2 - 1) as a coeff. , would simply y^5 as the coefficient still make the eq. linear? Also , for an equation to be non-linear , must the coefficients involve the dependent var. instead of the independent one?

Thank you very much!

 

[X89codered89X]

supposing y is "space" then, this is a linear, spatially-variant ODE. It wouldn't be considered nonlinear unless you couldn't write this ode as

...a_1(y) \frac{dx}{dy} + a_0(y) x = 0

If any of the "a" coefficients had an x or any of its derivatives in it, it would be nonlinear.

For example the following cannot be written in the above form (and are non linear)

...(y^2 - x)\frac{dx}{dy} + a_0(y) x = 0

...(y^2 - 1)(\frac{dx}{dy})^2 + a_0(y) x = 0

...(x^2 - 1)\frac{dx}{dy} + x^2 = 0

But the last example is at least "spatially invariant".

 

[HallsofIvy]

The equation you give is linear because It is "linear" in the dependent variable x. Specifically, if we had two functions, x_1(y) and x_2(y), that satisfy that equation, and any two numbers, \alpha and \beta, then the "linear combination", \alpha x_1(y)+ \beta x_2(y) also satisfies it:
a_1(y)\frac{d(\alpha x_1+ \beta x_2}{dy}+ a_0(y)(\alpha x_1+ \beta x_2)
= \alpha\left(a_1(y)\frac{d x_1}{dy}+ a_2(y)x_1\right)+ \beta \left(a_1(y)\frac{dx_2}{dy}+ a_2(y)x_2\right)= \alpha(0)+ \beta(0)= 0