<s>[INST] Solve the following Diffrential Equation $(3y-7x+7)dx-(3x-7y-3)dy=0$ I want to solve the following equation:
$$(3y-7x+7)dx-(3x-7y-3)dy=0$$
I will need  two new variables? or I can solve it with 1, for example set expression as $z$?
What you are suggesting?thanks.
 [/INST] Answer taken from  
http://in.answers.yahoo.com/question/index?qid=20080830232129AAdNlmN 
$$(3x−7y−3)dy=(3y−7x+7)dx \\
\frac{dy}{dx}=\frac{3y−7x+7}{3x−7y−3} $$
Now, let $x=x’+h$ and $y=y’+k$  
Therefore
$$\frac{dy’}{dx’} =\frac{3y’−7x’+3k−7h+7}{3x’−7y’+3h−7k−3} $$
Putting $h=1$ and $k=0$, we get  
$$\frac{dy’}{dx’}=\frac{3y’−7x’}{3x’−7y’} $$
Now, let $y’=vx’$ so that $$\frac{dy’}{dx’}=v+x’\frac{dv}{dx’}$$  
We have
$$ v+x’ \frac {dv}{dx’} = \frac {3vx’−7x’}{3x’−7vx’} $$  
$$x’ \frac {dv}{dx’} = \frac {3v−7}{3−7v} − v $$  
$$x’ \frac {dv}{dx’} = \frac {3v−7−3v+7v²}{3−7v} $$  
$$x’ \frac {dv}{dx’} = \frac {7v²−7}{3−7v} $$  
$$ \frac{dx’}{x’} = \frac {3−7v}{7v²−7} dv $$  
Integrating both sides,
$$\int\frac{dx’}{x’}=\int\frac{3 dv}{7v²−7}−\int\frac{7v dv}{7v²−7} \\
\int\frac{dx’}{x’}=\frac{3}{7} \int\frac{dv}{v²−1}−\int\frac{v dv}{v²−1} \\
ln|x’|+c=\frac{3}{7} \frac{1}{2} ln\frac{|v−1|}{|v+1|} 
−\frac{1}{2} \int\frac{v dv}{v²−1} \\
ln|x’|+c=\frac{3}{14} ln\frac{|v−1|}{|v+1|}
−\frac{1}{2} ln|v²−1| \\
ln|x’|+c=\frac{3}{14} ln\frac{|y’−x’|}{|y’+x’|}
−\frac{1}{2} ln|\frac{(y’)^2−(x’)^2}{(x’)^2}| $$  
(since $v=\frac{y’}{x’}$)  
$$ ln |x’|+c=\frac{3}{14} ln|y’−x’|−\frac{3}{14} ln|y’+x’|−
\frac{1}{2} ln|y’+x’|−\frac{1}{2} ln|y’−x’|
+\frac{1}{2} ln|(x’)^2| \\
ln |x’|+c=−\frac{2}{7} ln|y’−x’|−\frac{5}{7} ln|y’+x’|+\frac{1}{2} ln|(x’)^2| \\
2ln|y’−x’|+5ln|y’+x’|=C \\
2ln|y−x+1|+5ln|y+x−1|=C  $$
(since $y’=y$ and $x’=(x−1)$
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