Solving ODE: 2y(1+x^2√y)dx + x(2+x^2√y)dy = 0 | Step-by-Step Guide

  • Thread starter Thread starter AdrianZ
  • Start date Start date
  • Tags Tags
    Ode
AdrianZ
Messages
318
Reaction score
0

Homework Statement


2y(1+x^2\sqrt{y})dx + x(2+x^2\sqrt{y})dy = 0

The Attempt at a Solution


well, I substituted x^2√y=u but then when I tried to differentiate it I understood it would be so hard. Please check and see if I've differentiated it correctly:

√y = u/x^2 -> y = u^2.x^-4 -> dy/dx = 2u.u'x^(-4) - 4x^(-5).u^2
Is that correct? if yes, then I think I've just made the problem harder. how can I solve that ODE?
 
Physics news on Phys.org
It is correct. Substitute for y and y' in the original equation, simplify and arrange: it is a separable ODE.

ehild
 
You could try grouping the terms into perfect differentials. For example, 2 y dx + 2 x dy = 2 d(xy). The other two terms can be written similarly as d(something) . Then write u = xy and v = something and it is a simple linear ODE.
 
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