# Thread: 1st order linear equation

1. ## 1st 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 .

2. 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?

3. It is the standard substitution for homogeneus equations i.e. equations of the form

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

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

Fernando Revilla

Edited: Sorry, I didn't see Ackbeet's post.

4. Originally Posted by MathsLion
(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 .
The idea behind that substitution is to transform the equation into another one that's separable. can you solve it now?

5. ## Initial 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

6. 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?

7. Originally Posted by MathsLion
(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 .
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.