[SOLVED] Question 3

**Altair** · March 4th 2008, 04:49 PM

The answer for $ x sin (\frac {y}{x})y' = xcos(\frac {y}{x}) + ysin(\frac {y}{x}) $ is given as,

$ xcos(\frac {y}{x}) = c. $

Where as my answer after simplifying the following step,

$ \frac {dv}{cot(v)} = \frac {dx}{x} $

was,

$ -cos (\frac {y}{x}) = xc $

I have tried my best to get the required form but I couldn't. Where does the problem lie?

**mr fantastic** · March 5th 2008, 03:43 AM

Originally Posted by Altair

The answer for $ x sin (\frac {y}{x})y' = xcos(\frac {y}{x}) + ysin(\frac {y}{x}) $ is given as,

$ xcos(\frac {y}{x}) = c. $

Where as my answer after simplifying the following step,

$ \frac {dv}{cot(v)} = \frac {dx}{x} $

was,

$ -cos (\frac {y}{x}) = xc $

I have tried my best to get the required form but I couldn't. Where does the problem lie?

$\frac {dv}{cot(v)} = \frac {dx}{x} \Rightarrow \tan v \, dv = \frac{1}{x} \, dx \Rightarrow \int \tan v \, dv = \int \frac{1}{x} \, dx$

$\Rightarrow -\ln |\cos v| = \ln |x| + A \Rightarrow \ln |1/\cos v| = \ln |x| + A \Rightarrow \frac{1}{\cos v} = Bx$ where $B = e^A$

$\Rightarrow \frac{1}{B} = x \cos v \Rightarrow C = x \cos \left( \frac{y}{x} \right)$ , where C = 1/B.

**mr fantastic** · March 5th 2008, 03:55 AM

Originally Posted by Altair

The answer for $ x sin (\frac {y}{x})y' = xcos(\frac {y}{x}) + ysin(\frac {y}{x}) $ is given as,

$ xcos(\frac {y}{x}) = c. $

Where as my answer after simplifying the following step,

$ \frac {dv}{cot(v)} = \frac {dx}{x} $

was,

$ -cos (\frac {y}{x}) = xc $

I have tried my best to get the required form but I couldn't. Where does the problem lie?

The mistake you probably made was in not realising that $- \ln a = \ln a^{-1} = \ln \frac{1}{a}$ ......

**Altair** · March 5th 2008, 04:10 AM

Exactly.

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