For equation:

$$ 2x\tan(y)dx + \left(x^2 - 2\sin(y)\right)dy = 0 $$

in a book, a computation towards obtaining integrating factor is done, without showing intermediate steps:

$$ B(y) = - \frac{\frac{\partial P}{\partial y} - \frac{\partial Q}{\partial x}}{P} = -\tan(y) $$

$P$ is '$2x \tan(y)$', $Q$ is '$x^2 - 2\sin(y)$'. When I do the computation I obtain:

$$ -\frac{\frac{2x}{\cos^2y} - 2x}{2x \tan(y)} $$

Which is way away from simple $-\tan(y)$. Is the book wrong, or maybe there is a way to obtain the simple $-\tan(y)$ ?

PS. Well, the mine computation can be apparently written as:

$$ \frac{\frac{1-\cos ^2 y}{\cos ^2 y}}{\tan y} $$

what gives $\sin^2 y : \cos^2 y : \tan y = \tan y$, wonder why WolframAlpha doesn't reduce this that way (outputs sophisticated equation: $\left(1-\cos ^2 (y)\right) \csc(y) \sec (y)$), maybe there is a reason for this?

  • $\begingroup$ $\displaystyle \frac{1}{\cos^2 y}=1+\tan^2 y$. Does it help? $\endgroup$ – Galc127 Mar 5 '16 at 17:29


What you obtained is the integrating factor $-\tan(y)$. This is not the solution of the ODE. Your job isn't finished yet. So, do not compare to the WolframAlpha result which is the completed work.

$$-\tan(y)\left( 2x\tan(y)dx+(x^2+2\sin(y))dy \right)=0$$ So, you have still to integrate $2x\tan^2(y)$ relatively to $x$ only and the second one $\tan(y)\left(x^2+2\sin(y)\right)$ relatively to $y$ only.

The first one gives $x^2\tan^2(y)+f(y)$

Continue with the second one in order to determine $f(y)$

After having obtained $f(y)$ :

$d\left( x^2\tan^2(y)+f(y) \right)=0\quad \to \quad x^2\tan^2(y)+f(y)=$constant.

This allows to express : $x(y)$ and more complicated: $y(x)$.

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