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))
