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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.

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2 Answers 2

up vote 3 down vote accepted

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 $c$.

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@Berci Thank you, not sure why I wrote $2$! –  Tom Oldfield Apr 6 '13 at 23:04
    
:) $\,\,\!\!\!\,$ –  Berci Apr 6 '13 at 23:04
1  
@TomOldfield because the derivative of the product of two function is equal to zero that means that it is constant? –  Daniel Lopez Apr 6 '13 at 23:07
    
Understand it never mind, thanks. I will mark yours as answer both answers were really good anyway. –  Daniel Lopez Apr 6 '13 at 23:08
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@DanielLopez yes, it is true for a general function $f$ that $f' = 0$ means that $f$ is constant. (whether they are a product of $2$ other functions or not.) –  Tom Oldfield Apr 6 '13 at 23:09

About your confusion: this is just a converted version of the original differential equation: $$(x^2y)'=0\,.$$ And the other hint is that we have $f'=0 \implies f=$ constant.

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Thanks both answer are really nice. –  Daniel Lopez Apr 6 '13 at 23:08

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