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

##### 1 Answer

# ln| (y+1)/(y-1) | -y = x^2 + C #

#### Explanation:

We have:

# (y^2+1)dy/dx+2xy^2=2x # ..... [A]

We can rearrange this non-linear First Order differential equation [A] as follows:

# (y^2+1)dy/dx = 2x - 2xy^2#

# :. (y^2+1)dy/dx = -2x(y^2-1)#

# :. (y^2+1)/(y^2-1)dy/dx = -2x#

This is now separable, so we can "seperate the variables" to get:

# int \ (y^2+1)/(y^2-1) \ dy = int \ -2x \ dx# ..... [B]

The RHS integral is standard, and the LHS will require a little manipulation, as follows:

# int \ (t^2+1)/(t^2-1) \ dt = int \ (t^2-1+2)/(t^2-1) \ dt #

# " " = int \ 1 + 2/(t^2-1) \ dt #

# " " = int \ 1 + 2/((t+1)(t-1)) \ dt #

We can now decompose the fractional part of the integrand into partial fractions, as follows:

# 2/((t+1)(t-1)) -= A/(t+1) + B/(t-1) #

# " " = ( A(t-1) + B(t+1) ) / ( (t+1)(t-1) ) #

Leading to the identity:

# 2 -= A(t-1) + B(t+1) #

Where

Put

# t = -1 => 2 = -2A => A = -1 #

Put# t = +1 => 2 = +2B => B = +1 #

So using partial fraction decomposition we have:

# int \ (t^2+1)/(t^2-1) \ dt = int \ 1 - 1/(t+1) + 1/(t-1) \ dt #

Using this result we can now integrate [B] as follows:

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

# :. int \ -1 + 1/(y+1) - 1/(y-1) \ dy = int \ 2x \ dx #

# :. - y + ln| y+1 | - ln| y-1 | = x^2 + C #

# :. ln| (y+1)/(y-1) | -y = x^2 + C #

Which, is the **General Solution** .

We are unable to find a particular solution, as requested, as noi initial conditions have been provided to allow the constant