(x+y)dx=(-x)dy

I know this equation requires a substitution of v=y/x at some point, but I cannot see how am supposed to use this. Can anyone help please .

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- Jan 12th 2011, 10:35 AMMathsLion1st order linear equation
(x+y)dx=(-x)dy

I know this equation requires a substitution of v=y/x at some point, but I cannot see how am supposed to use this. Can anyone help please . - Jan 12th 2011, 10:48 AMAckbeet
This is a homogeneous equation. That is, M and N are both homogeneous functions of the same degree. Hence, the substitution you mentioned will render the equation separable. You have y = vx. Then dy = v dx + x dv. Substitute in and simplify. What do you get?

- Jan 12th 2011, 10:48 AMFernandoRevilla
It is the standard substitution for homogeneus equations i.e. equations of the form

$\displaystyle P(x,y)dx+Q(x,y)dy=0$

where $\displaystyle P$ and $\displaystyle Q$ are homogeneus functions of degree $\displaystyle d$ (in this case, $\displaystyle d=1$ ).

Fernando Revilla

Edited: Sorry, I didn't see**Ackbeet**'s post. - Jan 12th 2011, 11:20 AMwonderboy1953
- Jan 12th 2011, 11:21 AMMathsLionInitial Condition
Thank you very much.

I solved the equation using this method, and obtained log(x)+C=y/x.

I then checked this by using the attached file.

However, the problem I am faced with is with that of the initial condition.

If the initial condition is y(x=0)=-2, my constant keeps vanishing and do not know how to write my answer in EXPLICIT form.

Thank you - Jan 12th 2011, 11:24 AMAckbeet
You're trying to apply the initial condition at a point at which the solution is invalid. Are you sure that's the correct initial condition?

- Jan 12th 2011, 12:02 PMmr fantastic
You're expected to learn something from your other threads: http://www.mathhelpforum.com/math-he...de-166809.html. If you did not understand what was told to you over there, you should have said so at that thread.