(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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(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
dx+Q(x,y)dy=0)
whereand
are homogeneus functions of degree
(in this case,
).
Fernando Revilla
Edited: Sorry, I didn't see Ackbeet's post.
The idea behind that substitution is to transform the equation into another one that's separable. can you solve it now?Quote:
Originally Posted by MathsLion
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 trying to apply the initial condition at a point at which the solution is invalid. Are you sure that's the correct initial condition?
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.Quote:
Originally Posted by MathsLion