Limits and DEs

Use separation of variables to integrate for $x$ : $\frac {dy}{dx} = - \frac {x\,(1+y^{2})^{2}}{\ln(y) \,(1 - y^{2})}$ $\therefore \int x\,\mathrm{d}x = - \int \ln(y)\,\frac {1-y^{2}}{(1+y^{2})^{2}} \,\mathrm{d}y$ $\frac {1}{2}x^{2} = - \int \ln(y)\,\frac {1-y^{2}}{(1+y^{2})^{2}} \,\mathrm{d}y$ Now use integration by parts to evaluate the right-hand-side where $f'(y) = \frac {1-y^{2}}{(1+y^{2})^{2}}$ , $g(y) = \ln(y)$ and hence $g'(y) = \frac {1}{y}$ . To find $f(y)$ use partial fractions where $z=y^{2}$ : $\frac {1-z}{(1+z)^{2}} = \frac {A}{(1+z)^{2}} + \frac {B}{1+z}$ $1-z = {A} + B\,(1+z)$ When $z = -1$ , $A = 2$ , hence $B = -1$ for all $z$ . Rewriting the expression for $f(y)$ , now use the substitution $u = \arctan y$ and $\mathrm{d}y = \sec^{2}u\,\mathrm{d}u$ for the integral on the left side. $f(y) = \int \frac {2}{(1+y^{2})^{2}}\,\mathrm{d}y - \int \frac {1}{1+y^{2}}\,\mathrm{d}y$ $f(y) = \int \frac {2}{(1+\tan^{2}u)^{2}\,\sec^{2}(u)}\,\mathrm{d}u - \arctan y$ $f(y) = \int 2\cos^{2}(u)\,\mathrm{d}u - \arctan y$ Use the identity $\cos 2u = 2\cos^{2}(u) - 1$ to simplify: $f(y) = \int \cos 2u +1 \,\mathrm{d}u - \arctan y$ $= \frac {1}{2} \sin 2u + u - \arctan y$ $= \sin u \cos u + u - \arctan y$ $= \sin (\arctan y) \cos (\arctan y) + \arctan y - \arctan y$ Use the identities $\sin (\arctan y) = \frac {y}{\sqrt{1+y^{2}}}$ and $\cos (\arctan y) = \frac {1}{\sqrt{1+y^{2}}}$ to obtain $f(y) = \frac {y}{1+y^{2}}$ . Now we can go back to the original integral using integration by parts. $\frac {1}{2}x^{2} = - \int \ln(y)\,\frac {1-y^{2}}{(1+y^{2})^{2}} \,\mathrm{d}y$ $\frac {1}{2}x^{2} = - \frac {y}{1+y^{2}}\,\ln y + \int \frac {1}{y}.\frac {y}{1+y^{2}} \,\mathrm{d}y$ $\frac {1}{2}x^{2} = - \frac {y}{1+y^{2}}\,\ln y + \arctan y + c$ Substitute the point (1, 1) to obtain the constant of integration: $\frac {1}{2} = \frac {\pi}{4} + c$ $c = \frac {1}{2} - \frac {\pi}{4}$ $\frac {1}{2}x^{2} = -\frac {y}{1+y^{2}}\,\ln y + \arctan y + \frac {1}{2} - \frac {\pi}{4}$ $x = \pm \sqrt{-\frac {2y}{1+y^{2}}\,\ln y + 2\arctan y + 1 - \frac {\pi}{2}}$ To satisfy the condition $y(1)=1$ , the negative result is ignored. $x = \sqrt{-\frac {2y}{1+y^{2}}\,\ln y + 2\arctan y + 1 - \frac {\pi}{2}}$ $\lim_{y \to \infty} x(y) = \lim_{y \to \infty} \sqrt{-\frac {2y}{1+y^{2}}\,\ln y + 2\arctan y + 1 - \frac {\pi}{2}}$ As $y \to \infty$ , $-\frac {2y}{1+y^{2}}\,\ln y \to 0$ and $2\arctan y \to \pi$ . Hence the limit becomes: $\sqrt{0 + \pi + 1 - \frac {\pi}{2}} = \sqrt{\frac {\pi}{2} + 1}$ $\therefore A + B + C = \boxed{4}$