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) #
