8(b)

With $$M = x^2 - y^2$$ and N = -2 y, we have

$$\frac{\partial M/\partial y - \partial N/\partial x}{N} = \frac{-2y}{-2y} = 1$$

This does not depend on y, so there is an integrating factor $\mu(x)$that depends only on x, with $\mu' = \mu$. A solution of this is $\mu(x) = \exp(x)$.

Now the solution is F(x,y) = C where $$\partial F/\partial x = \mu M = {\rm e}^x (x^2- y^2)$$ and $$\partial F/\partial y = \mu N = -2 {\rm e}^x y$$. Integrating $\mu M$ with respect to x, we have

$$F(x,y) = \int {\rm e}^x (x^2 - y^2)\, dx + h(y) = {\rm e}^x (x^2 - 2 x + 2 - y^2) + h(y)$$

Differentiating with respect to y, $$\partial F/\partial y = -2y {\rm e}^x + h'(y) = \mu N = -2 y e^x$$ so h'(y) = 0 and we can take h(y) = 0. Thus the general solution is $${\rm e}^x (x^2 - 2 x + 2 - y^2) = C$$. From the initial value, $C = {\rm e}^2$. We can solve for y explicitly: $$y = \pm \sqrt{x^2 - 2 x + 2 - {\rm e}^{2-x}}$$. Since y(2) = 1 and not -1 we must use +, not -:

$$y = \sqrt{x^2 - 2 x + 2 - {\rm e}^{2-x}}$$