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March 5, 2015

March 5, 2015

Posted by **Taldil** on Tuesday, February 26, 2013 at 6:09am.

- Maths -
**Steve**, Tuesday, February 26, 2013 at 10:16amThis is just a special case of the Bernoulli Equation

y' + p(x) y = q(x) y^n

where q(x) = 1

Make the substitution v = y^(1-n)

This will give you

-v' + (1+1/x)v = 1

Now you can multiply by an integrating factor

e^[∫(1+1/x) dx] = xe^x to get

xe^x v = -e^x (x-1)

and you're almost done

- Maths -
**Taldil**, Wednesday, February 27, 2013 at 6:43amHi..

I'm not sure how to get the final answer

**Answer this Question**

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