<s>[INST] differential linear equation of order one $(2xy+x^2+x^4)\,dx-(1+x^2)\,dy=0$ I have no idea how to solve it. Should be linear equation of order one since I am passing through this chapter, but I can't put into the form of $$y'+P(x)y=Q(x)$$
Here is the equation: $$(2xy+x^2+x^4)\,dx-(1+x^2)\,dy=0$$
It is not exact since partial derivatives are not equal.
Any help would be appreciated.
 [/INST] $$(2xy+x^2+x^4)\,dx−(1+x^2)\,dy=0\\
(1+x^2) \frac {dy}{dx} = (2xy+x^2+x^4)\\
y' - \frac {2x}{1+x^2} y = x^2$$
Integrating factor
$$e^{\int \frac {-2x}{1+x^2}} = e^{-\ln(1+x^2)} = \frac {1} {(1+x^2)}$$
$$\frac {1}{1+x^2}y = \int \frac {x^2}{1+x^2}\\
\frac {1}{1+x^2}y = x - \arctan x + C\\
y = x^3 + x - (1+x^2)\arctan x + C(1+x^2)$$
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