Perform partial fractions decomposition
#1/((1+x)(3+x)) = A/(1+x)+B/(3+x)#
#1/((1+x)(3+x)) = (A(3+x)+B(1+x))/((1+x)(3+x))#
#1/((1+x)(3+x)) = ( (A+B)x +(3A+B) ) / ( (1+x) (3+x) )#
#{(A+B=0),(3A+B=1):}#
#{(A=-B),(-3B+B=1):}#
#{(A=1/2),(B=-1/2):}#
So:
#1/((1+x)(3+x)) = 1/(2(1+x))-1/(2(3+x))#
Solve now:
#dy/dx = y/((1+x)(3+x)) #
Separate the variables:
#dy/y = dx/((1+x)(3+x))#
#lnabsy = int dx/((1+x)(3+x)) = 1/2 int dx/(1+x) -1/2int dx/(3+x)+C#
#lnabsy = 1/2 ln abs(1+x) -1/2ln abs(3+x) +C#
For #x > -1# both arguments are positive, so:
#lnabsy = 1/2 ln (1+x) -1/2ln (3+x) = ln sqrt((1+x)/(3+x))+C#
Taking the exponential:
#absy = c_1 sqrt((1+x)/(3+x))#
where #c_1=e^C# is positive. Admitting also negative value for #c_1# and noting that #y(x) = 0# is also a solution, then the general solution is:
#y(x) = c sqrt((1+x)/(3+x))# with #c in RR#
We can determine the particular solution for:
#y(1) = 2#
#c sqrt((1+1)/(3+1)) = 2#
#c sqrt(1/4) = 2#
#c/2= 2#
#c=4#
Finally:
#y(x) = 4sqrt((1+x)/(3+x))#
