Question

y' = (2y+1)/(x^2+x). Is this a separable differential equations and how should i solve it?

Answer 1

Go to this site and see if it helps a bit. www.wolframalpha.com

Answer 2

And yes, I believe they are separable differential equations.

Answer 3

The "x belongs to R" is strange because x is usually a variable.

First solve y" - y' - 2y = 0. This has associated equation p^2 - p - 2 = 0 which has solutions p = 2 or -1. General solution is thus y = A*e^2x + B*e^-x.

Particular solutions usually have the same form as RHS so assume y = ax + b, y' = a, y" = 0
y" - y' - 2y = 0 - a - 2(ax + b) = x
This leads to a = -1/2 and b = 1/4

Together these give y = A*e^2x + B*e^-x - x/2 + 1/4.

x = 0 ----> y = A + B + 1/4 = 1
x = 0 ----> y' = 2A - B - 1/2 = -1
These lead to A = 1/12 and B = 2/3

Final answer y = (1/12)*e^2x + (2/3)*e^-x - x/2 + 1/4
Does that help.

Answer 4

\[y'=\frac{2y+1}{x^2+x}=\frac{2y+1}{x(x+1)}\]\[\frac{y'}{2y+1}=\frac{1}{x(x+1)}\]\[\frac{1}{2}\left(\ln{(2y+1)}\right)'=\frac{1}{x}-\frac{1}{x+1}\]\[\frac{1}{2}\ln{(2y+1)}=\ln{x}-\ln{(x+1)}+C=\ln{\frac{C_1x}{x+1}}\]\[\ln{(2y+1)}=\ln{\left(\frac{C_1x}{x+1}\right)^2}\]\[y=\frac{1}{2}\left[\left(\frac{C_1x}{x+1}\right)^2-1\right]\]