Is This the Correct Approach to Solving the Exact Differential Equation?

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Homework Statement


2(y^2+1)dx+(4xy-3y^2)dy=0

Homework Equations

The Attempt at a Solution


here is my attempt to solve
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That's one way of solving it (observing that it is exact). I didn't check all steps though, so don't know if you are correct. You can fill in your solution in the differential equation and see if it works.

An alternative approach would be to divide both sides by ##x^2## and substitute ##z= y/x##
 
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Math_QED said:
That's one way of solving it (observing that it is exact). I didn't check all steps though, so don't know if you are correct. You can fill in your solution in the differential equation and see if it works.

An alternative approach would be to divide both sides by ##x^2## and substitute ##z= y/x##
im sorry but there is no z in the problem , i think you've mistaken 2 for z sorry about my penmanship :) lol
 
Baconslider said:
im sorry but there is no z in the problem , i think you've mistaken 2 for z sorry about my penmanship :) lol

No, you are misunderstanding me.

You can divide both sides by ##x^2## and you will get an expression in ##y/x##. You can consider ##y/x## as a new variable, which we will call ##z##. Thus, ##z = y/x## and ##xz = y##, so ##dy = dx z + xdz## and you can use this to transform your differential equation to one in ##z(x)##. Here, it is unnecessary as the DE is exact, but sometimes the method I just explained can be useful :)
 
Baconslider said:

Homework Statement


2(y^2+1)dx+(4xy-3y^2)dy=0

Homework Equations


The result you got is not correct. I can not follow what you did, but M and N are functions of both x and y. When you integrate M(x,y) with respect to x or N(x,y) with respect to y, you get two x,y functions. I do not understand, what you F(x) and F(y) are. Also, why is g'(x)=1, and if it so, why is g(x)=0? You should type in your solution. It would be easier to check what you wrote even for yourself.
 
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...
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