# Solving First Order Linear Differential Equation

1. Apr 22, 2012

### the0

1. The problem statement, all variables and given/known data

Given (E): $(x+1)^{2}(xy'-y) = -(2x+1)$

Determine the set of applications from the interval $I$ to $ℝ$ which are solutions of (E) for:

a) $I = (0,+∞)$
b) $I = (-1,0)$
c) $I = (-∞,-1)$
d) $I = (-1,+∞)$
e) $I = ℝ$

The attempt at a solution

I have rearranged (E) into the form:

$y'-\frac{y}{x}$ = $\frac{-(2x+1)}{x(x+1)^{2}}$

and chosen integrating factor:

$exp(∫\frac{-1}{x}dx) = \frac{1}{x}$

now multiplying by the integrating factor and integrating gives:

$∫\frac{y'}{x}-\frac{y}{x^{2}}dx = ∫\frac{-(2x+1)}{x^{2}(x+1)^{2}}dx$

partial fractions now give:

$∫\frac{y'}{x}-\frac{y}{x^{2}}dx = ∫\frac{-1}{x^{2}}+\frac{1}{(x+1)^{2}}dx$

After integrating and a tiny bit of algebra I get a general solution, for $k\inℝ$ as:

$y=\frac{1}{x+1}+kx$

Now I think I am right in saying that this solution works fine for the intervals in a), b) and c)?

However for $I = (-1,+∞)$ there is (or potentially is) a problem with the point $0$ and for $I = ℝ$ there are (or potentially are) problems with the points $0$ and $-1$?

Now I'm not sure how to proceed. Am I thinking along the right lines?

Any help/pointers would be very much appreciated, Thanks!

2. Apr 23, 2012

### the0

Can the general solution already posted simply extend to $I = (-1,+∞)$ ??
If so how can this be shown in detail?

Also I don't yet understand what happens when we are considering $ℝ$

Thanks for any help!