# Thread: A really simple integration...

1. ## A really simple integration...

Find the particular solution for dy/dx=(y-5)/x, using the point y(0)=7

I got y-5=Ax (or ln(y-5)=ln(x)+c)

For both of which there is no way to find A or c...I had this question in a test, and wrote it was impossible to find the particular solution with the given point (I believe the examiner made an error..) am I correct?

For both of which there is no way to find A or c...I had this question in a test, and wrote it was impossible to find the particular solution with the given point (I believe the examiner made an error..) am I correct?
You are wrong, you need to use y(0)=7 to find A or c.

Do you follow?

3. Yea...and when you sub x=0 into ln(x) you get an error...and 7-5 doesnt = 0*A...so....

4. The DE can be written as...

$\displaystyle x y' - y = - 5$ (1)

... so that it is a linear first order DE. Neccessary condition to have a solution is that the coefficient of the term y' doesn't vanish... but the 'initial condition' is for $\displaystyle x=0$ so that that isn't verified...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

5. ... but if we swap the x and y variables the DE becomes...

$\displaystyle x'= \frac{x}{y-5}$ , $\displaystyle x(7)=0$ (1)

... the solution of which is $\displaystyle x=0$... very easy ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$