An integral of a differential equation that's troubling me

Question

I am facing a problem in differential equations.

$$(x-x^3)dy = (y+yx^2-3x^4)dx\tag{Question}$$

I am completely recognisant that I can use the linear differential equation form here, as I have shown below:

$$\frac{dy}{dx}=\frac{y+yx^2-3x^4}{x-x^3}=y\frac{1+x^2}{x-x^3}-\frac{3x^3}{1-x^2}$$

$$\frac{dy}{dx}-y\frac{1+x^2}{x-x^3}=-\frac{3x^3}{1-x^2}$$

From the above LDE form, I can deduce the below

$$Integrating\space factor = e^{-\int\frac{(1+x^2)}{x-x^3} \,dx}\tag{Problematic one}$$

Now I tried catching WolframAlpha (I did not know how to proceed, but eventually I recognized the partial fraction present), and it gave me the answer for the integrating factor as $$\frac{1-x^2}{x}$$ This would further complicate things as the final solution would look like:

$$y\times\frac{1-x^2}{x}=-\int\frac{3x^3}{1-x^2}\times\frac{1-x^2}{x}dx\tag{Problematic one}$$

which makes it an excessively lengthy problem.

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This problem is from the JEE Mains 2021 where barely 3 minutes is given for a problem and this would be quite solvable if and only if Physics and Chemistry were over in an hour or so.

I am a 12th grader where we learn only substitution, partial fractions and integration by parts here.

Any suggestions for alternative methods to do this faster?

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EDIT: After the complete error rectification now the problem seems fine. I will be able to proceed.

Answer 1

We are given

$$(x-x^3)dy = (y-yx^2-3x^4)dx$$

We can write this as

$$\dfrac{dy}{dx}=\dfrac{y-yx^2-3x^4}{x-x^3}=y\dfrac{1-x^2}{x-x^3}-\dfrac{3x^3}{1-x^2}$$

You seem to have a simple algebra issue, it is $1-x^2$, not $1+x^2$.

You should have gotten $$\dfrac{1-x^2}{x-x^3} = \dfrac{1}{x}$$

The integral and exponential are easy now.

Answer 2

Rewrite the equation as $$ \frac{d y}{d x}+\frac{y\left(1+x^2\right)}{x\left(x^2-1\right)}=\frac{3 x^3}{x^2-1} \tag*{(*)} $$ Then the integrating factor is $$ e^{\int \frac{2 x+1}{x(x+1)} d x} =e^{\ln \left(\frac{1-x^2}{x}\right)} =\frac{1-x^2}{x} $$ Multiplying the equation $(*)$ by $\frac{1-x^2}{x}$ yields $$ \begin{aligned} \frac{1-x^2}{x} \frac{d y}{d x}-\frac{\left(1+x^2\right) y}{x^2}&=-3 x^2 \\ \frac{d}{d x}\left(\frac{1-x^2}{x} y\right)&=-3 x^2 \\ \frac{1-x^2}{x} y&=-x^3+C \end{aligned} $$ Hence the solution of the differential equation is $$ y=\frac{Cx}{1-x^2}+\frac{x^4}{x^2-1} $$ where $C$ is a constant.