1. ## [SOLVED] Question 3

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

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

Where as my answer after simplifying the following step,

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

was,

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

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

2. Originally Posted by Altair
The answer for $\displaystyle x sin (\frac {y}{x})y' = xcos(\frac {y}{x}) + ysin(\frac {y}{x})$ is given as,

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

Where as my answer after simplifying the following step,

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

was,

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

I have tried my best to get the required form but I couldn't. Where does the problem lie?
$\displaystyle \frac {dv}{cot(v)} = \frac {dx}{x} \Rightarrow \tan v \, dv = \frac{1}{x} \, dx \Rightarrow \int \tan v \, dv = \int \frac{1}{x} \, dx$

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

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

3. Originally Posted by Altair
The answer for $\displaystyle x sin (\frac {y}{x})y' = xcos(\frac {y}{x}) + ysin(\frac {y}{x})$ is given as,

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

Where as my answer after simplifying the following step,

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

was,

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

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 $\displaystyle - \ln a = \ln a^{-1} = \ln \frac{1}{a}$ ......

4. Exactly.