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Suppose you have the differential equation $$y\,dx+y\,dy=0 \tag 1\label1,$$ where $(y)$ is a non zero function.

The solution is $(x_2-x_1)+(y_2-y_1)=0 \tag 2\label2.$$

If you divide the original equation by $(y)$ or $(y^2)$, etc., you will always have the same solution.

Now, suppose you have a condition, say: $$y=\frac c x\quad \text{($xy$ a constant)}\tag 3\label3.$$

The original equation becomes $$\frac c x dx+y \,dy=0\tag 4\label4.$$

The solution is: $$c \ln\frac{x_2}{x_1}+(y_2-y_1)=0 \tag 5\label5.$$

Note that the condition $\eqref 3$ should be inserted in the equation and taken into account when solving, i.e. if you use this condition with the solution $\eqref 2$, which was derived without the condition inserted in the equation, you will not have the right result.

Now, suppose you divide equation $\eqref 4$ with $(y^2)$, i.e. with $\bigl(\frac c x\bigr)^2$. The new equation you have is $$\frac x c+\frac{dy}y=0\tag 6\label6.$$

The solution is $$\frac{x_2^2-x_1^2}{2c} + \ln \left(\frac{y_2}{y_1}\right) \tag7\label7$$

Unlike the case of dividing the equation $\eqref1$ where by $y^2$ (or other power of $y$) we always have the same solution, in equation $\eqref 6$ the solution is different.

So, why are the solutions $\eqref5$ and $\eqref7$ different?

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I do not understand why you let $y = c/x$, but then only choose to use that in one place. –  Amzoti Nov 5 '13 at 3:59
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