#1/((1+x)(3+x)) = alpha/ (1+x) + beta/(3+x) #
#= (alpha(3+x)+ beta(1+x))/((1+x)(3+x)) = 1/((1+x)(3+x)) #
Look at numerators and let #x = -1 #
#implies 2 alpha = 1, alpha = 1/2#
Look at numerators and let #x = -3 #
#implies -2 beta = 1, beta = -1/2#
#implies 1/((1+x)(3+x)) = 1/(2 (1+x)) - 1/(2(3+x)) #
Your DE is now:
#y' = y (1/(2 (1+x)) - 1/(2(3+x)))#
This separates as:
#(2y')/ y = 1/( (1+x)) - 1/((3+x))#
We integrate wrt x:
#2 int 1/ y y' dx = int 1/( (1+x)) - 1/((3+x)) dx#
By chain rule:
#= 2 int 1/ y dy = int 1/( (1+x)) - 1/((3+x)) dx#
#implies 2 ln y = ln (1+x) - ln(3+x) + C#
With #C# as a generic constant, ie the actual letter is oblivious to operations:
#implies ln y = ln sqrt ((1+x)/(3+x)) + C#
Put both sides to power of #e#:
#e^(ln y) = e^( ln sqrt ((1+x)/(3+x)) + C)#
#y = e^( ln sqrt ((1+x)/(3+x))) e^ C = C sqrt ((1+x)/(3+x))#
#y(1) = 2 implies 2 = C cdot sqrt(2/4) implies C = 2 sqrt 2#
#implies y = 2 sqrt2 sqrt ((1+x)/(3+x))#
