- #1
the0
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Homework Statement
Given (E): [itex](x+1)^{2}(xy'-y) = -(2x+1)[/itex]
Determine the set of applications from the interval [itex]I[/itex] to [itex]ℝ[/itex] which are solutions of (E) for:
a) [itex]I = (0,+∞)[/itex]
b) [itex]I = (-1,0)[/itex]
c) [itex]I = (-∞,-1)[/itex]
d) [itex]I = (-1,+∞)[/itex]
e) [itex]I = ℝ[/itex]
The attempt at a solution
I have rearranged (E) into the form:
[itex]y'-\frac{y}{x}[/itex] = [itex]\frac{-(2x+1)}{x(x+1)^{2}}[/itex]
and chosen integrating factor:
[itex]exp(∫\frac{-1}{x}dx) = \frac{1}{x}[/itex]
now multiplying by the integrating factor and integrating gives:
[itex]∫\frac{y'}{x}-\frac{y}{x^{2}}dx = ∫\frac{-(2x+1)}{x^{2}(x+1)^{2}}dx[/itex]
partial fractions now give:
[itex]∫\frac{y'}{x}-\frac{y}{x^{2}}dx = ∫\frac{-1}{x^{2}}+\frac{1}{(x+1)^{2}}dx[/itex]
After integrating and a tiny bit of algebra I get a general solution, for [itex]k\inℝ[/itex] as:
[itex]y=\frac{1}{x+1}+kx[/itex]
Now I think I am right in saying that this solution works fine for the intervals in a), b) and c)?
However for [itex]I = (-1,+∞)[/itex] there is (or potentially is) a problem with the point [itex]0[/itex] and for [itex]I = ℝ[/itex] there are (or potentially are) problems with the points [itex]0[/itex] and [itex]-1[/itex]?
Now I'm not sure how to proceed. Am I thinking along the right lines?
Any help/pointers would be very much appreciated, Thanks!