Non-Linear Differential.

$\dfrac{dx}{dt} = a^2 x +a^2 y - b^2 x ( x^2 + y^2)$

$\dfrac{dy}{dt} = -a^2 x + a^2 y -b^2 y(x^2 + y^2)$

Converting to polar coordinates, let $x = r \cos(\theta), y = r \sin(\theta)$ $x^2 + y^2 = r^2$ and $\theta = \arctan(\dfrac{y}{x})$ $$

Taking the derivatives with respect to $t$ :

$x^2 + y^2 = r^2 \implies r \dfrac{dr}{dt} = x \dfrac{dx}{dt} + y \dfrac{dy}{dt}$

and

$\theta = \arctan(\dfrac{y}{x}) \implies r^2 \dfrac{d\theta}{dt} = x \dfrac{dy}{dt} - y \dfrac{dx}{dt}$ $$

$(1): x * (\frac{dx}{dt} = a^2 x +a^2 y - b^2 x ( x^2 + y^2))$

$(2): y * (\frac{dy}{dt} = -a^2 x + a^2 y -b^2 y(x^2 + y^2))$

and adding $(1)$ and $(2)$ we obtain:

$r \dfrac{dr}{dt} = r^2 (a^2 - b^2 r^2) \implies \dfrac{dr}{dt} = r (a^2 - b^2 r^2)$

$(3): -y * (\frac{dx}{dt} = a^2 x +a^2 y - b^2 x ( x^2 - y^2))$

$(4): x * (\frac{dy}{dt} = -a^2 x + a^2 y -b^2 (x^2 - y^2))$

and adding $(3)$ and $4$ we obtain:

$\dfrac{d\theta}{dt} = -a^2$ $$

The system of differentials above has a single critical point at $r = 0$ . For $r > 0$ , the system above reduces to:

$\dfrac{dr}{dt} = r (a^2 - b^2 r^2)$

$\dfrac{d\theta}{dt} = -a^2$

$\implies \dfrac{1}{a^2} \int (\dfrac{1}{r} + \dfrac{b}{2 (a - b r)} - \dfrac{b}{2 (a + b r)}) dr = \dfrac{1}{a^2} \ln(\dfrac{r}{\sqrt{a^2 - b^2 r^2}}) = t + C \implies$ $$

$\dfrac{r^2}{a^2 - b^2 r^2} = C e^{2 a^2 t} \implies$ $r = \dfrac{a}{\sqrt{b^2 + C e^{-2 a^2 t}}}$ and $\theta = -a^2 t + t_{0}$ . $$

Using initial conditions $x(t = 0) = y(t = 0) = \dfrac{1}{2}$ , where $r(0) = \sqrt{x(0)^2 + y(0)^2} \implies \dfrac{a}{\sqrt{b^2 + C}} = \dfrac{1}{\sqrt{2}} \implies C = 2 a^2 - b^2$

Let $\beta = \frac{1}{a^2} (\ln(\frac{\sqrt{2 a^2 - b^2}}{b}))$ and $\alpha = \frac{1}{a^2} (\ln(\frac{\sqrt{3(2 a^2 - b^2)}}{b}))$

For $I = \displaystyle\int_{\beta}^{\alpha} r(t) dt =$ $a\displaystyle\int_{\beta}^{\alpha} \dfrac{e^{a^2 t}}{\sqrt{2a^2 - b^2 + (be^{a^2 t})^2}}$ $$

Let $b e^{a^2 t} = \sqrt{2 a^2 - b^2} \tan\theta \implies$ $dt = \dfrac{\sec^2(\theta)}{a^2 \tan\theta}d\theta$

and for $\alpha$ we have $\tan(\theta) = \sqrt{3}$ and for $\beta$ we have $\tan(\theta) = 1 \implies$

$I = \dfrac{1}{ab} \displaystyle\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \sec\theta \: d\theta =$ $$

$\dfrac{1}{ab} \ln(\sec\theta + \tan\theta)|_{\frac{\pi}{4}}^{\frac{\pi}{3}} =$ $$

$\dfrac{1}{ab} \ln(\dfrac{2+ \sqrt{3}}{\sqrt{2} + 1}) =$ $\dfrac{1}{ab} \ln \left(\dfrac{\alpha + \sqrt{\beta}}{\sqrt{\alpha} + \lambda}\right)$ $\implies \alpha + \beta + \lambda = \boxed{6}$