# Thread: first order DE show that

1. ## first order DE show that

show that the general solution of

dy/dx=exp(px-qy) , with p,q E Reals, can be written as y= aln(bexp(px)+w)

and express a, b and w in terms of p,q and an arbritrary constant C

I believe I have to integrate as differentiating y will not produce a constant so

let px-qy=v

dy/dx=(p-dv/dx)/q

dv/dx=p-qe^v=p(1-(q/p)e^v)

seperate variables

int(p/p-qe^v)dv=int(p)dx

v-ln(p-qe^v)+c=px

qy= c-ln(p-qexp(px-qy))

How to get in required form?

2. Originally Posted by poirot
show that the general solution of

dy/dx=exp(px-qy) , with p,q E Reals, can be written as y= aln(bexp(px)+w)

and express a, b and w in terms of p,q and an arbritrary constant C

I believe I have to integrate as differentiating y will not produce a constant so

let px-qy=v

dy/dx=(p-dv/dx)/q

dv/dx=p-qe^v=p(1-(q/p)e^v)

seperate variables

int(p/p-qe^v)dv=int(p)dx

v-ln(p-qe^v)+c=px

qy= c-ln(p-qexp(px-qy))

How to get in required form?
I haven't looked at your substitution method in any detail, but it might work. However I think you are overthinking this. The equation separates nicely in its original form.

-Dan

3. Which equation are you referring to ? I was thing about the law ln(ab)=ln(a) +ln(b) but unfortunatly it is a sum, not a product that is being logged.

4. Originally Posted by poirot
dy/dx=exp(px-qy) , with p,q E Reals, can be written as y= aln(bexp(px)+w)
$\displaystyle \frac{dy}{dx} = e^{px - qy}$

$\frac{dy}{dx} = e^{px}e^{-qy}$

$e^{qy}dy = e^{px}dx$

-Dan

5. Yes I realised last night.
After integration, I get 1/q(e^qy)+c=1/pe^px,

y=1/q(ln(q/p(e^px)-qc))

w= -qc (which I know is another constant but I thought it best to be consistent)
a=1/q, b=q/p