dy/dx= (x+1)y/2x^2 where y(1)=1, for all x>=1
i ended up with
y^2/2= 1/2 ln (2x^2)+1/4 tan-1 (x/2) + K
anythoughts?
I think this DE is separable
$\displaystyle \frac{dy}{dx}= \frac{(x+1)y}{2x^2}$
$\displaystyle \frac{dy}{y}= \frac{(x+1)}{2x^2}~dx$
$\displaystyle \int\frac{1}{y}~dy= \int \frac{(x+1)}{2x^2}~dx$
$\displaystyle \int\frac{1}{y}~dy= \int \frac{x}{2x^2}+\frac{1}{2x^2}~dx$
$\displaystyle \int\frac{1}{y}~dy= \int \frac{1}{2x}+\frac{1}{2x^2}~dx$
Did you use this method?
I think that solution is better, from here raise each side of the equation to the power of $\displaystyle e$ . This will eliminate the $\displaystyle \ln(y)$ on the left hand side.
Before this, may I ask how you arrived at the $\displaystyle \tan^{-1}(x)$ term? I can't see this being correct.
I see, I'm not sure if you have applied that the correct way here.
consider
$\displaystyle \int\frac{1}{y}~dy= \int \frac{1}{2x}+\frac{1}{2x^2}~dx$
$\displaystyle \ln(y)= \int \frac{1}{2x}~dx+\int \frac{x^{-2}}{2}~dx$
$\displaystyle \ln(y)= \frac{\ln(x)}{2}- \frac{x^{-1}}{2}+C$