# How do I solve this differential equation? # (x^2 + y^2) dx + xydy = 0#

##### 1 Answer

The GS is:

# y^2 = (A -x^4)/(2x^2) #

Or, alternatively:

# y = +-sqrt(A -x^4)/(sqrt(2)x) #

#### Explanation:

We have:

# (x^2 + y^2) \ dx + xy \ dy = 0 #

Which we can write in standard form as:

# dy/dx = -(x^2 + y^2)/(xy) # ..... [1]

Which is a non-separable First Order Ordinary Differential Equation. A suggestive substitution would be to perform a substitution of the form:

# y = xv => dy/dx = v + xv' \ \ \ # where#v=v(x) #

Then if we substitute into the DE [1], we get

# v + (dv)/dx = -(x^2 + (xv)^2)/(xvx) #

# :. v + x(dv)/dx = -(1 + v^2)/(v) #

# :. x(dv)/dx = -(1 + v^2)/(v) -v #

# :. x(dv)/dx = -(1 + v^2)/(v) -v^2/v #

# :. x(dv)/dx = -(1 + 2v^2)/(v) #

# :. v/(1+2v^2) (dv)/dx = -1/x #

Which has transformed the initial DE [1] into a separable, DE, so we can Manipulate further, and **"separate the variables"**:

# 1/4 \ int \ (4v)/(1+2v^2) \ dv =- \ int \ 1/x \ dx #

And we can now integrate to get:

# 1/4 ln |1+2v^2| =- ln|x| + C #

And, noting that

# ln (1+2v^2) =- 4ln|x| + 4C #

# :. ln (1+2v^2) + 4ln|x| - lnA = 0 \ \ \ # , say

# :. ln (1+2v^2) + lnx^4 - lnA = 0 #

# :. ln (x^4/A(1+2v^2)) = 0 #

# :. x^4/A(1+2v^2) = e^0 #

And restoring the substitution, we can now write:

# :. x^4/A(1+2(y/x)^2) = 1 #

# :. 1+2(y/x)^2 = A/(x^4) #

# :. 2(y/x)^2 = A/(x^4) -1 #

# :. 2y^2/x^2 = (A -x^4)/x^4 #

# :. y^2 = (A -x^4)/(2x^2) #

Or, alternatively:

# y = +-sqrt(A -x^4)/(sqrt(2)x) #