# 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: