What is the general solution of the differential equation # dy/dx=(y+2x)^2+2 #?
1 Answer
Sep 27, 2017
# y = 2tan(2x + A) - 2x #
Explanation:
We have:
# dy/dx=(y+2x)^2+2 # ..... [A]
We can perform a substitution, Let
# v = y + 2x #
Then if we differentiate, we get:
# (dv)/dx = dy/dx + 2 => dy/dx = (dv)/dx - 2 #
Substituting into the original DE
# (dv)/dx - 2 = v^2+2 #
# :. (dv)/dx = v^2+4 #
This is now a First Order separable DE, so we can rearrange and seperate the variables, giving:
# int \ 1/(v^2+4) \ dv = int \ dx #
Which we can integrate, using standard integrals, to get:
# 1/2arctan (v/2) = x + C #
And, restoring the substitution:
# 1/2arctan ((y + 2x)/2) = x + C #
# arctan ((y + 2x)/2) = 2x + 2C #
# :. (y + 2x)/2 = tan(2x + A) #
# :. y+2x = 2tan(2x + A) #
# :. y = 2tan(2x + A) - 2x #
