# (x+y-1)dx+(y-x-5)dy = 0

1. May 20, 2013

### skyturnred

1. The problem statement, all variables and given/known data

Solve (x+y-1)dx+(y-x-5)dy=0

2. Relevant equations

3. The attempt at a solution

x=u+h
y=v+k

Therefore

h=-2 and k=3

Therefore

(u+v)du+(v-u)dv=0

$\frac{dv}{du}$=$\frac{u+v}{u-v}$

This is where I'm stuck. I can't solve the equation above. I have a feeling I need to rearrange so that I have a function of $\frac{v}{u}$ on the right hand side, but I am unable to rearrange it to attain this.

Any suggestions?

Thank-you

2. May 20, 2013

### CAF123

The purpose of the initial substitution was to find an appropriate h,k such that the constant terms in the eqn in the OP vanished and hence arrive at precisely the form you now have. Define another variable $f = \frac{v}{u}$ after rearranging the RHS of your eqn.

3. May 20, 2013

### skyturnred

OK, I have tried this already. I still get stuck however.

g=$\frac{v}{u}$

v=gu

$\frac{dv}{du}$=u$\frac{dg}{du}$+g

so when you sub this into $\frac{dv}{du}$=$\frac{u+v}{u-v}$ you get

u$\frac{dg}{du}$+g=$\frac{\frac{v}{g}+gu}{\frac{v}{g}-gu}$

And I'm still stuck. This seemed even more complicated than before the substitution.

4. May 20, 2013

### CAF123

You have the eqn $$\frac{dv}{du} = \frac{u+v}{u-v} = \frac{1 + \frac{v}{u}}{1 - \frac{v}{u}}$$ You correctly computed dv/du to obtain $u\frac{dg}{du} + g$ but the RHS of the eqn that you wrote above is incorrect. You want a RHS to be entirely in terms of g and u so you can apply separation of variables.

5. May 20, 2013

### skyturnred

Thank-you. I will be able to solve it from here on.

My biggest issue was in realizing that $\frac{u+v}{u-v}$=$\frac{1+\frac{v}{u}}{1-\frac{v}{u}}$

After this I rearranged to

$\frac{du}{u}$=$\frac{1-g}{1+g^{2}}$dg

I should be able to solve this using techniques of integration.

Thank-you once again