What is the general solution of the differential equation #(x+y)dx-xdy = 0#?

Question

Isaiah Allen · Accepted Answer

# y = xln|x| + Ax #

We can write the equation # (x+y)dx-xdy = 0 # as:

# " " xdy = (x+y)dx # # :. dy/dx = (x+y)/x # # :. dy/dx = 1+y/x #

# :. dy/dx - 1/x y = 1 #

This is a First Order DE of the form:

# y'(x) + P(x)y = Q(x) #

Which we know how to solve using an Integrating Factor given by:

# IF = e^(int P(x) \ dx) #

And so our Integrating Factor is:

# IF = e^(int -1/x \ dx) # # \ \ \ \ = e^(-ln|x|) # # \ \ \ \ = e^(ln|1/x|) # # \ \ \ \ = 1/x #

If we multiply by this Integrating Factor we will (by its very design) have the perfect differential of a product:

# " " dy/dx - 1/x y = 1 #

# :. 1/xdy/dx - 1/x^2 y = 1/x # # :. \ \ \ \ \ d/dx(1/xy) = 1/x #

Which is now a separable DE, and we can separate the variables to get:

# " "y/x = int \ 1/x \ dx # # :. y/x = ln|x| + A # (where #A# is an arbitrary constant) # :. \ \ y = xln|x| + Ax #

Emma Aldridge · Answer

undefined To find the general solution of the given differential equation, we first rearrange it as follows:

(x + y) dx - x dy = 0
(x + y) dx = x dy

Now, we divide both sides by x(x + y):

(dx)/(x) = (dy)/(x + y)

Integrating both sides:

∫(dx/x) = ∫(dy/(x + y))

ln|x| = ln|x + y| + C

where C is the constant of integration.

Exponentiating both sides:

|x| = e^(ln|x + y| + C)
|x| = e^(ln|x + y|) * e^C
|x| = |x + y| * e^C

Since e^C is just another constant, let k = e^C. Thus,

|x| = k|x + y|

This can be separated into two cases:

1. If x ≠ 0, then |x| = k(x + y) or x = ±k(x + y)

2. If x = 0, then |0| = k(0 + y) or 0 = 0 (no new solution)

Hence, the general solution is:

x = ky - ky, where k is any nonzero constant.