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Quiz 1 / Re: Q1, P1 Night sections

« on: October 04, 2013, 07:26:42 AM »

I will show my solution using the second method given in the hint.

First, we will find $z$.

$a(t)z' + b(t)z = 0$
$\Rightarrow t^2z' + 2tz = 0$
$\Rightarrow (t^2z)' = 0$
$\Rightarrow t^2z = c$, where $c$ is an arbitrary constant
$\Rightarrow z = \frac{c}{t^2}$

I will choose $c$ to be 1 and substitute $y = zu = \frac{1}{t^2}u$. Then $\frac{dy}{dt} = -\frac{2}{t^3}u + \frac{1}{t^2}\frac{du}{dt}$.
From the Bernoulli equation,

$t^2y' + 2ty - y^3 = 0$
$\Rightarrow t^2(-\frac{2}{t^3}u + \frac{1}{t^2}\frac{du}{dt}) + 2t(\frac{1}{t^2}u) - (\frac{1}{t^2}u)^3 = 0$
$\Rightarrow -\frac{2u}{t} + \frac{du}{dt} + \frac{2u}{t} - \frac{u^3}{t^6} = 0$
$\Rightarrow \frac{du}{dt} = \frac{u^3}{t^6}$
$\Rightarrow \frac{1}{u^3}du = \frac{1}{t^6}dt$

The last equation above is separable. Integrating the left side with respect to $u$ and the right side with respect to $t$ gives us

$-\frac{1}{2u^2} = -\frac{1}{5t^5} + k$, where $k$ is an arbitrary constant
$\Rightarrow 1 = \frac{2u^2}{5t^5} - 2ku^2$
$\Rightarrow 1 = \frac{2}{5t}y^2 - 2ky^2t^4$
$\Rightarrow 1 = y^2(\frac{2}{5t} - 2kt^4)$
$\Rightarrow 1 = y^2(\frac{2 - 10kt^5}{5t})$
$\Rightarrow y^2 = \frac{5t}{2 - 10kt^5}$
$\Rightarrow y = \pm\sqrt{\frac{5t}{2 - 10kt^5}}$