Basic Math Problem with Differential Equations: Integrating Factors

Question

Please can anyone break down in detailed steps how these 2 lines are equal to each other. I'm confused.

Also please ignore the " = 0 " in the second line.

\begin{align*} (1+x^2)\,&\frac{dy}{dx}+\frac{d(1+x^2)}{dx}\cdot y \\ &\frac{d}{dx}\,(y(1+x^2))=0 \end{align*}

Thanks a lot in advance.

Answer 1

The idea here is using the product rule backwards. To see things clearly take $a = 1+ x^2$ and keep $y$ as it is.

We know that from product rule,

$\frac{d(ay)}{dx} = y \frac{da}{dx} + a \frac{dy}{dx}$

here the equality is used backwards,

$ y \frac{da}{dx} + a \frac{dy}{dx} = \frac{d(ay)}{dx}$

If you are new to product rule and wonder about how it is true, start thinking from the first principles of derivative definition or look at https://en.wikipedia.org/wiki/Product_rule.

Answer 2

$$ (1+x^2)\frac{dy}{dx}+\frac{d(1+x^2)}{dx}\cdot y =\frac{d}{dx}\,(y(1+x^2)) $$ Differentiate the right side

You differentiate a product of two functions, just apply the rule $$\frac{d}{dx}(f \times g)= g\frac{d}{dx}f+f\frac{d}{dx}g$$ $$\frac{d}{dx}\,(y(1+x^2))=(1+x^2)\frac {d}{dx}y+y\frac {d}{dx}(x^2+1)$$ $$\frac{d}{dx}\,(y(1+x^2))=(1+x^2)y'+2xy$$ note that $$\frac{d}{dx}\,(1+x^2)=2x$$