dy/dx - y/(x+1) = x y(1)= 0, x>0
integral of [Ae^-ln(x+1)(x)]/[Ae^-ln(x+1)]
y= x/(x+1) - ln(x+1)/(x+1) + c/(x+1)
0= 1/2 -(ln2)/2 + C/3
y= x/(x+1) -ln(x+1)/(x+1) + (-3+ln2)/2(x+1)
is the answer right?
No.
$\displaystyle \frac{dy}{dx} - \frac{1}{x + 1}y = x$
The integrating factor is $\displaystyle e^{\int{-\frac{1}{x+1}\,dx}} = e^{-\ln{(x + 1)}} = e^{\ln{(x + 1)^{-1}}} = (x + 1)^{-1}$.
So multiply both sides of the equation by $\displaystyle (x + 1)^{-1}$ and you get
$\displaystyle (x + 1)^{-1}\frac{dy}{dx} - (x + 1)^{-2}y = \frac{x}{x + 1}$
$\displaystyle \frac{d}{dx}[(x + 1)^{-1}y] = 1 - \frac{1}{x + 1}$.
Can you go from here?
Let's see...
$\displaystyle \frac{d}{dx}[(x + 1)^{-1}y = 1 - \frac{1}{x + 1}$
$\displaystyle (x + 1)^{-1}y = x - \ln{|x + 1|} + C$
$\displaystyle y = (x + 1)(x - \ln{|x + 1|} + C)$.
We're told $\displaystyle y(1) = 0$ so
$\displaystyle 0 = (1 + 1)(1 - \ln{|1 + 1|} + C)$
$\displaystyle 0 = 2(1 - \ln{2} + C)$
$\displaystyle 0 = 1 - \ln{2} + C$
$\displaystyle C = \ln{2} - 1$.
So $\displaystyle y = (x + 1)(x - \ln{|x + 1|} + \ln{2} - 1)$.