Q:

Need help in solving this differential equation, understand hint given but still clueless.

Issue
Okay I know this differential equation is just plain simple, but I still cannot figure it out even when there is a hint attached next to it. I understand what the hint is--really saying but somehow I am missing the bigger picture in order to solve it. It is like I need one more hint or something to figure it out. I even solved the other odd problems and got it right but this one just haunts me. I am no math expert but I am not that bad in math at least in my opinion.
The Problem
Here is the differential equation problem, I will even include the hint that was written nicely for me.
$$x^2y'+2xy = 0$$
$$y(1) = 2$$
Hint: Interpert the left-hand side of the equation as the derivative of a product of two functions.
How I Understand The Hint
Okay the hint is saying that
$$x^2y'+2xy$$ is the same as
$$(x^2y)'$$
Correct if I am wrong with my understanding of the hint, because maybe this is why I am not getting it right.
The Confusion
So am I supposed to equate the derivate of two products with the left hand side as shown below?
$$(x^2y)'=x^2y'+2xy$$
Because I tried the separating variables technique but got a natural logarithm in my answer, which shows me that I am getting a banana when I am supposed to get an apple. Plus I do not think I can use that technique because it is taught until the next chapter.
Conclusion
All I need is some guidance, and by the way this is not homework, not taking a Calculus class this semester but just practicing some problem to exercise my math skills.

A:

You are correct in your interpretation of the hint, it tells us that: $x^2y'+2xy=0$ is equivalent to saying that $(x^2y)' = 0$. This second equation is easier to solve, because integrating both sides with respect to $x$ gives that $x^2y=c$ for some constant $c$, and so $y = \frac{c}{x^2}$. We can then use the given initial condition to find the value of $c$.

