(a) The differential equation is separable, so we separate the variables and integrate:
[tex](x+1)\dfrac{dy}{dx} = xy \implies \dfrac{dy}y = \dfrac x{x+1} \, dx = \left(1-\dfrac1{x+1}\right) \, dx[/tex]
[tex]\displaystyle \frac{dy}y = \int \left(1-\frac1{x+1}\right) \, dx[/tex]
[tex]\ln|y| = x - \ln|x+1| + C[/tex]
When x = 0, we have y = 2, so we solve for the constant C :
[tex]\ln|2| = 0 - \ln|0 + 1| + C \implies C = \ln(2)[/tex]
Then the particular solution to the DE is
[tex]\ln|y| = x - \ln|x+1| + \ln(2)[/tex]
We can go on to solve explicitly for y in terms of x :
[tex]e^{\ln|y|} = e^{x - \ln|x+1| + \ln(2)} \implies \boxed{y = \dfrac{2e^x}{x+1}}[/tex]
(b) The curves y = x² and y = 2x - x² intersect for
[tex]x^2 = 2x - x^2 \implies 2x^2 - 2x = 2x (x - 1) = 0 \implies x = 0 \text{ or } x = 1[/tex]
and the bounded region is the set
[tex]\left\{(x,y) ~:~ 0 \le x \le 1 \text{ and } x^2 \le y \le 2x - x^2\right\}[/tex]
The area of this region is
[tex]\displaystyle \int_0^1 ((2x-x^2)-x^2) \, dx = 2 \int_0^1 (x-x^2) \, dx = 2 \left(\frac{x^2}2 - \frac{x^3}3\right)\bigg|_0^1 = 2\left(\frac12 - \frac13\right) = \boxed{\frac13}[/tex]
