What is the general solution of the differential equation? # (x+1)y'+y= tan^-1(x) / (x+1) #
1 Answer
# (1+x)y = int \ tan^-1(x) /(1+x) \ dx + C #
Explanation:
We have:
# (x+1)y'+y= tan^-1(x) / (x+1) #
We can re-arrange this ODE as follows:
# dy/dx + 1/(x+1) y = tan^-1(x)/(1+x)^2 # ..... [1]
This is a First Order Linear non-homogeneous Ordinary Differential Equation of the form;
# dy/dx + P(x)y=Q(x) #
We can readily generate an integrating factor when we have an equation of this form, given by;
# I = e^(int P(x) dx) #
# \ \ = exp(int \ 1/(1+x) \ dx) #
# \ \ = exp( ln(1+x) ) #
# \ \ = (1+x) #
And if we multiply the DE [1] by this Integrating Factor,
# (1+x)dy/dx + y = tan^-1(x) /(1+x) #
# d/dx((1+x)y) = tan^-1(x) /(1+x) #
(Which is incidentally the equation we started with). We can directly integrate to get:
# (1+x)y = int \ tan^-1(x) /(1+x) \ dx + C #
And in terms of elementary functions this is as far we can get:
