Toronto Math Forum

$$y'' + \frac{x}{x^2}y' + \frac{x^2-v^2}{x^2}y = 0$$
$$W = ce^{-\int p(t)dt}$$
$$p(t) = \frac{1}{x}$$
$$-\int p(t)dt = -\ln{x}$$
$$W = ce^{-\ln{x}} = ce^{\ln{\frac{1}{x}}} = \frac{c}{x}$$

$$y′′−2y′−2y=0$$
$$r^2−2r−2=0$$
$$r = \frac{−b \pm \sqrt{b^2−4ac}}{2a}$$
$$r_1=1+\sqrt{3}, r_2=1−\sqrt{3}$$

$$y=c_1e^{(1+\sqrt{3})t}+c_2e^{(1-\sqrt{3})t}$$

$p(t) = -1, q(t) = 1+3 \sin(t)$

$\mu(t) = e^{\int p(t)dt} = e^{\int (-1)dt}=e^{-t}$

$\mu y' - \mu y = \mu (1+3\sin(t))$

$e^{-t} y' - e^{-t}y = e^{-t} (1+3\sin(t))$

$\frac{d}{dt}(e^{-t} y) = e^{-t} +3e^{-t}\sin(t)$

$\int \frac{d}{dt}(e^{-t} y) dt= \int e^{-t} +3e^{-t}\sin(t) dt$

$e^{-t} y = -e^{-t} - \frac{3}{2}\sin te^{-t} - \frac{3}{2}\cos te^{-t} + c$

$y(t) = -1 - \frac{3}{2}\sin t - \frac{3}{2}\cos t + ce^t$

$y(0) = y_0 = -1 - \frac{3}{2}\sin(0) - \frac{3}{2}\cos(0) + ce^0 = -1 - \frac{3}{2} + c$

$c = y_0 + \frac{5}{2}$

$y(t) = -1 - \frac{3}{2}\sin t - \frac{3}{2}\cos t + (y_0 + \frac{5}{2}) e^t$

If $y_0 < - \frac{5}{2}, y_0 + \frac{5}{2} < 0$,

$$\lim_{t\to\infty} y(t) =  \lim_{t\to\infty} (y_0 + \frac{5}{2}) e^t = -\infty$$
If $y_0 > - \frac{5}{2}, y_0 + \frac{5}{2} > 0$,
$$\lim_{t\to\infty} y(t) =  \lim_{t\to\infty} (y_0 + \frac{5}{2}) e^t = +\infty$$
If $y_0 = - \frac{5}{2}, y_0 + \frac{5}{2} = 0$,
$$\lim_{t\to\infty} y(t) =  \lim_{t\to\infty} (-1 - \frac{3}{2}\sin t - \frac{3}{2}\cos t)$$
is the only value of $y_0$ that makes the solution finite
$y_0 = - \frac{5}{2}$