(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 .
(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 .
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?
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.
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
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.