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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?2012-11-07
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    @MartinSleziak, What is the difference between ode and differential-equations?2012-11-07
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    Most tags have tag-excerpt (which is displayed when you hover over the tag with you mouse) and [tag-wiki](http://math.stackexchange.com/tags/differential-equations/info). As you can read there, the [tag:differential-equations] tag should be used for [ordinary differential equations](http://en.wikipedia.org/wiki/Ordinary_differential_equation). I thought the correct tag was [tag:ode], which I remembered incorrectly. That's why I've edited your post twice.2012-11-07
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    @MartinSleziak I wasn't complaining, very happy with your edit, much better than changing ( to \left(.2012-11-07

1 Answers 1

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

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    what do you mean integrating factor?2012-11-07
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    http://en.wikipedia.org/wiki/Integrating_factor#Use_in_solving_first_order_linear_ordinary_differential_equations2012-11-07
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    I found an answer with wolfram alpha. But it needed to use integration.2012-11-07
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    If you're going to solve a differential equation, there's no way to avoid integration (definite or indefinite) at some stage...2012-11-07