<s>[INST] How to solve this 1st order ODE $$ \frac{dy}{dx} = \frac{2x^2 + 3y^2 - 7}{3x^2 + 4y^2 + 8}$$
This does not satisfy the exactness and I can't find any integrating factor to transform it. I can't make it homogeneous too.
Thanks in advance for the replies.
edit: The question was written incorrectly by the instructor, so in this version there is no analytic solution for it. 
 [/INST] Follow the method in http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=164:
Let $u=\dfrac{y}{x}$ ,
Then $y=xu$
$\dfrac{dy}{dx}=x\dfrac{du}{dx}+u$
$\therefore x\dfrac{du}{dx}+u=\dfrac{2x^2+3(xu)^2-7}{3x^2+4(xu)^2+8}$
$x\dfrac{du}{dx}=\dfrac{(3u^2+2)x^2-7}{(4u^2+3)x^2+8}-u$
$x\dfrac{du}{dx}=-\dfrac{(4u^3-3u^2+3u-2)x^2+8u+7}{(4u^2+3)x^2+8}$
$((4u^3-3u^2+3u-2)x^2+8u+7)\dfrac{dx}{du}=-(4u^2+3)x^3-8x$
Let $v=\dfrac{1}{x^2}$ ,
Then $\dfrac{dv}{du}=-\dfrac{2}{x^3}\dfrac{dx}{du}$
$\therefore-\dfrac{((4u^3-3u^2+3u-2)x^2+8u+7)x^3}{2}\dfrac{dv}{du}=-(4u^2+3)x^3-8x$
$\left(\dfrac{8u+7}{x^2}+4u^3-3u^2+3u-2\right)\dfrac{dv}{du}=\dfrac{8}{x^4}+\dfrac{4u^2+3}{x^2}$
$((8u+7)v+4u^3-3u^2+3u-2)\dfrac{dv}{du}=8v^2+(4u^2+3)v$
This belongs to an Abel equation of the second kind.
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