(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
where and are homogeneus functions of degree (in this case, ).
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.