What is the general solution of the differential equation # dy/dx + y^2=4#?
1 Answer
# y = 2(e^(4x)-16)/(e^(4x)+16)#
Explanation:
We have:
# dy/dx + y^2=4 #
We can rearrange this Differential Equation as follows:
# dy/dx =4 -y^2 #
# :. 1/(4-y^2) dy/dx = 1 #
This is a First Order separable Differential Equation, so we can "seperate the variables" to get:
# int \ 1/( 2^2-y^2) \ dy = int \ dx # ..... [A]
Using the standard result:
# int \ 1/(a^2-t^2) \ dt = 1/(2a) ln abs( (a+t)/(a-t) ) #
we can integrate [A], to get:
# 1/4 ln abs( (2+y)/(2-y) ) = x + C_1 #
# :. ln abs( (2+y)/(2-y) ) = 4x + C #
Using the initial condition,
# -ln abs( (2+0)/(2-0) ) = 4ln2 + C #
# :. 0 = 4ln2 + C #
# :. C = -4ln2 = - ln16#
Thus, the (implicit) solution is:
# ln abs( (2+y)/(2-y) ) = 4x - ln16 #
We can rearrange to get an explicit solution:
# :. abs( (2+y)/(2-y) ) = e^(4x - ln16) #
Noting that
# (2+y)/(2-y) = e^(4x)e^(-ln16) #
# :. (2+y)/(2-y) = e^(4x)/16 #
For simplicity, Put
# (2+y)/(2-y) = A #
# :. 2+y = A(2-y) #
# :. 2+y = 2A-Ay #
# :. y+Ay = 2A-2 #
# :. y(A+1) = 2(A-1) #
# :. y = 2(A-1)/(A+1) #
Thus, the solution is:
# y = 2(e^(4x)/16-1)/(e^(4x)/16+1) #
# \ \ = 2(e^(4x)/16-1)/(e^(4x)/16+1) xx 16/16#
# \ \ = 2(e^(4x)-16)/(e^(4x)+16)#
