Ordinary Differential Equation by substitution.

[WREX88]

1. The problem statement, all variables and given/known data
dy/dx = (x+3y)/(3x+y)

I have to solve the given differential equation by using an appropriate substitution...


3. The attempt at a solution

I used algebra to make the equation (3x+y)dy - (x+3y)dx = 0

then made x = vy and dx = vdy + ydv.

then plugged in to get (3vy+y)dy - (vy+3y)(vdy+ydv)

and now I'm looking for a common factor to cancel and I'm stuck, any help? Am I even headed in the right direction...thanks.

[Mark44]

It doesn't seem you're heading in the right direction. Here's what I would try.

Starting with this equation,
(3x+y)dy = (x+3y)dx

integrate the left side with respect to y. Since you're integrating with respect to y, you'll end up with whatever you get plus an arbitrary function of x alone.
Then, integrate the right side with respect to x, similar to above. This time around you'll need to include an arbitrary function of y alone.

The left and right sides have to be equal for all pairs of (x, y) values, so you can equate the arbitrary functions of x or y on either side with what's on the opposite side.

This should give you the equation h(x, y) = 0.

To check your answer, take the derivative implicitly.

[HallsofIvy]

Looking at the symmetry of "3x+ y" and "x+ 3y", which has "average" value 2x+ 2y= 2(x+ y), let u= x+ y and v= x- y. Then x= (1/2)u+ (1/2)v and y= (1/2)u- (1/2)v, dx= (1/2)du+ (1/2)dv, dy= (1/2)du- (1/2) dv. Put those into the equation. you should wind up with udv= vdu.