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**Prove It** You've been given the solution already, but I'll go step by step...

1. $\displaystyle \frac{dy}{dx} = \frac{1 + y^2}{1 + x^2}$

$\displaystyle \frac{1}{1 + y^2}\,\frac{dy}{dx} = \frac{1}{1 + x^2}$

$\displaystyle \int{\frac{1}{1 + y^2}\,\frac{dy}{dx}\,dx} = \int{\frac{1}{1 + x^2}\,dx}$

$\displaystyle \int{\frac{1}{1 + y^2}\,dy}= \int{\frac{1}{1 + x^2}\,dx}$

$\displaystyle \arctan{y} + C_1 = \arctan{x} + C_2$

$\displaystyle \arctan{y} = \arctan{x} + C$.

2. $\displaystyle \frac{dy}{dx} - xy = x$

$\displaystyle \frac{dy}{dx} = xy + x$

$\displaystyle \frac{dy}{dx} = x(y + 1)$

$\displaystyle \frac{1}{y + 1}\,\frac{dy}{dx} = x$

$\displaystyle \int{\frac{1}{y + 1}\,\frac{dy}{dx}\,dx} = \int{x\,dx}$

$\displaystyle \int{\frac{1}{y + 1}\,dy} = \int{x\,dx}$

$\displaystyle \ln{|y + 1|} + C_1 = \frac{1}{2}x^2 + C_2$

$\displaystyle \ln{|y + 1|} = \frac{1}{2}x^2 + C$

$\displaystyle |y + 1| = e^{\frac{1}{2}x^2 + C}$

$\displaystyle |y + 1| = e^C e^{\frac{1}{2}x^2}$

$\displaystyle y + 1 = \pm e^C e^{\frac{1}{2}x^2}$

$\displaystyle y + 1 = Ae^{\frac{1}{2}x^2}$

$\displaystyle y = Ae^{\frac{1}{2}x^2} - 1$.