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.
 A: We are given:
$$\tag 1 \dfrac{dy}{dx} = \dfrac{3y-7x+7}{3x-7y-3}$$
It would really be nice if we could get rid of those constants, so lets use a trick to do that.
Let: $X = x + h \rightarrow dX = dx, Y = y + k \rightarrow dY = dy$
We substitute into $(1)$ and get:
$$\tag 2 \dfrac{dY}{dX} = \dfrac{3(y+k)-7(x+h)+7}{3(x+h)-7(y+k)-3}$$
What we want now is to get rid of those constants, so we have:
$$3k - 7h + 7 = 0 \\ -7k + 3h -3 = 0$$
This gives us: $ h = 1, k = 0$.
Our new systems now becomes:
$$\tag 3 \dfrac{dY}{dX} = \dfrac{-7X + 3Y}{3X - 7Y}$$
Let $Y = vX \ \rightarrow Y' = v + X v'$ and substitute into $(3)$, which is now a separable equation and we end up with:
$$\displaystyle \int \dfrac{7v - 3}{v^2 - 1}~ dv = -7 \int \dfrac{1}{X}~dX$$
I think you can take it from here.
Recall that once you solve this, there are two substitutions to get back to the solution.
A: Hint
Solve the system 
$$(3y-7x+7)=0\\-(3x-7y-3)=0$$
Assume you get the solution as $(x=\alpha,y=\beta)$.
Make $x=\bar x+\alpha$ and $y=\bar y+\beta$ and substitute in differential equation. After that make $\bar y=u\bar x$ and $\ d\bar y=u\ d \bar x +\bar x\ du$.
A: 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 $x’=(x−1)$
