# An integral of a differential equation that's troubling me [closed]

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.

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?

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

• That integral is handled by long division, then partial fractions if needed. Jan 25 at 2:04
• @SeanRoberson I am looking for alternatives, the method I followed is excessively lengthy. Long division, then again partial fractions only lengthens the problem Jan 25 at 2:06
• On correcting the error, I get the form now, so I'll upload my answer soon. Jan 25 at 2:36
• It's off-topic now, but the "problematic" integral can be simplified a bit by replacing $x^2=\dfrac{1+y}{1-y}$ to get$$\int\frac{dy}{y-y^3}$$Still some partial fractions to deal with, and you trade polynomial division for simplifying a nested rational expression. Jan 25 at 2:37
• @user170231 I getcha. Jan 25 at 2:39

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.

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.