# What is the general solution of the differential equation #y' - (2xy)/(x^2+1) = 1#?

##### 1 Answer

# y = (x^2+1) arctan(x) + C(x^2+1) #

#### Explanation:

We have:

# y' - (2xy)/(x^2+1) = 1 # ..... [A}

We can use an integrating factor when we have a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

# dy/dx + P(x)y=Q(x) #

As the equation is already in this form, then the integrating factor is given by;

# I = e^(int P(x) dx) #

# \ \ = exp(int \ -(2x)/(x^2+1) \ dx) #

# \ \ = exp(- int \ (2x)/(x^2+1) \ dx) #

# \ \ = exp( - ln(x^2+1) #

# \ \ = exp( ln (1/(x^2+1) ) ) #

# \ \ = 1/(x^2+1) #

And if we multiply the DE [A] by this Integrating Factor,

# (1/(x^2+1)) \ y' - (1/(x^2+1)) (2xy)/(x^2+1) = (1/(x^2+1)) 1 #

# :. d/dx( (1/(x^2+1)) \ y ) = 1/(x^2+1) #

Which we can directly integrate to get:

# (y)/(x^2+1) = int \ 1/(x^2+1) \ dx #

The RHS is a standard integral, so integrating we get:

# (y)/(x^2+1) = arctan(x) + C #

Leading to the General Solution:

# y = (x^2+1) arctan(x) + C(x^2+1)#