# 1st order ode

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• Dec 23rd 2010, 08:57 AM
MathsLion
1st order ode
I am trying to solve the following ODE in order to obtain a solution in explicit for:

xdy=(x+y)dx

WITH INITIAL CONDITION Y(X=0)=-2

i have tried exact equations and attempted substituion but no luck.

thanks
• Dec 23rd 2010, 09:01 AM
Ackbeet
The equation is linear in y. Have you tried the integrating factor method?
• Dec 23rd 2010, 09:08 AM
Ackbeet
Hmm. Come to think of it, the point x = 0 is a singularity of the DE. I'm not sure you can solve it as written. Are you sure the initial condition is applied at x = 0?

Incidentally, could you please reduce the size of your font? It's a bit loud!
• Dec 24th 2010, 02:01 AM
General
It is homogeneous also.
• Dec 24th 2010, 02:34 AM
mr fantastic
Quote:

Originally Posted by General
It is homogeneous also.

@OP: To expand, the DE can be written $\displaystyle \displaystyle \frac{dy}{dx} = 1 + \frac{y}{x}$ (provided $\displaystyle x \neq 0$). I suppose the given initial condition might be a limit, as in $\displaystyle \displaystyle \lim_{x \to 0} y = -2$ .... But that condition is false in fact false since $\displaystyle \displaystyle \lim_{x \to 0} y = 0$ ....
• Dec 24th 2010, 03:34 AM
FernandoRevilla
The substitution $\displaystyle y=vx$ ( homogeneous equation ) provides the general solution:

$\displaystyle y=x(C+\log |x|)$

Possibly there is a typo and the initial condition is $\displaystyle y(1)=-2$ , so the particular solution would be:

$\displaystyle y=x(-2+\log |x|)$

Fernando Revilla