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As I was studying differential equations, I got a question in my mind that we are taking out general solutions for a differential equation. Like for example: $$\frac{dy}{dx} = \frac{1}{(1+x³)} - \frac{3x²}{(1 + x²)}y$$ Solution: $$\Rightarrow \frac{dy}{dx} + \frac{3x²}{(1 + x²)} y = \frac{1}{(1+x³)}$$

Comparing it with $\frac{dy}{dx} + Py = O$, we get

$$P = \frac{3x²}{1+x³}$$

$$Q= \frac{1}{1 + x³}$$

Let’s figure out the integrating factor(I.F.) which is $e^{\int P\,dx}$

$$ \Rightarrow ~\text{I.F}~ = e^{\int \frac{3x²}{(1+x³)}\,dx}=e^{\ln(1+x³)}$$

$$\Rightarrow~\text{I.F.}~ = 1 + x³$$

Now, we can also rewrite the L.H.S as:

$$\frac{d(y \times I.F)}{dx}, $$

$$ \Rightarrow d(y \times (1 + x³)) dx = \frac{1}{(1 +x³)} \times (1 + x³)$$

Integrating both the sides w. r. t. $x$, we get,

$$\Rightarrow y \times ( 1 + x³) = x$$

$$\Rightarrow y = \frac{x}{(1 + x³)}$$

$$\Rightarrow y = \frac{x}{(1 + x³)} + C$$

But how do we get that differential equation (as given in the above question $\rightarrow \frac{dy}{dx} = \frac{1}{(1+x³)} – \frac{3x²}{(1 + x²)}y$) from our real life ?

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  • $\begingroup$ Could you check your source, the denominator should have $(1+x^3)$ in both places. Then $3x^2$ can also be identified as the inner derivative. $\endgroup$ – Lutz Lehmann Jan 13 at 19:56
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    $\begingroup$ This would be much easier to read if you used Mathjax. $\endgroup$ – DMcMor Jan 13 at 20:02
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    $\begingroup$ Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. $\endgroup$ – Shaun Jan 13 at 20:04
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I am afraid that, with the $x^2$ and $x^3$, there is no solution.

For the homogeneous equation, we have $$y= C \,e^{-3 \left(x-\tan ^{-1}(x)\right)}$$ Variation of the parameter leads to $$C'=\frac{e^{3 (x- \tan ^{-1}(x))}}{1+x^3}$$ which is impossible to integrate.

Same problem if we had $x^2$ twice.

So, as already said in comments, the problem must be $$\frac{dy}{dx} = \frac{1}{(1+x^3)} - \frac{3x²}{(1 + x^3)}y$$ which is very simple.

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