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I was told to solve this without integration and to use implicit diffentiation.

$$x^3 y^{\prime} - \dfrac{3y}{x} = x^3 e^{\left(x - \dfrac{1}{x^3}\right)}$$

I am utterly lost, any suggestions.

I can get to

$$ x^4y^{\prime} - 3y = x^4 e^{\left(x - \dfrac{1}{x^3}\right)}$$

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@martin sleziak, any ideas? – yiyi Nov 7 '12 at 6:52
@MartinSleziak, What is the difference between ode and differential-equations? – yiyi Nov 7 '12 at 6:56
Most tags have tag-excerpt (which is displayed when you hover over the tag with you mouse) and tag-wiki. As you can read there, the differential-equations tag should be used for ordinary differential equations. I thought the correct tag was ode, which I remembered incorrectly. That's why I've edited your post twice. – Martin Sleziak Nov 7 '12 at 7:00
@MartinSleziak I wasn't complaining, very happy with your edit, much better than changing ( to \left(. – yiyi Nov 7 '12 at 9:20
up vote 1 down vote accepted

Hint: Multiply both sides by the integrating factor $e^{1/x^3}$.

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what do you mean integrating factor? – yiyi Nov 7 '12 at 10:54… – Hans Lundmark Nov 7 '12 at 11:11
I found an answer with wolfram alpha. But it needed to use integration. – yiyi Nov 7 '12 at 12:29
If you're going to solve a differential equation, there's no way to avoid integration (definite or indefinite) at some stage... – Hans Lundmark Nov 7 '12 at 16:32

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