How can I solve this second order differential equation?

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Homework Statement



dx/dt=1-1/y
dy/dt=1/(x-t)

The Attempt at a Solution



If I take the derivative of the second equation and substitute it to the first one I

\frac{d^2 y}{dt^2} - \frac{1}{y} (\frac{dy}{dt})^2 = 0

but I don't know how to solve this. Could anyone name any methods that could try?
Thanks.
 
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The trick is to notice
<br /> y&#039;&#039;-\frac{1}{y}(y&#039;)^{2}=0\Rightarrow\frac{y&#039;&#039;}{y&#039;}=\frac{y&#039;}{y}<br />
 
<br /> y&#039;&#039;-\frac{1}{y}(y&#039;)^{2}=0\Rightarrow yy&#039;&#039;= y&#039;^{2}

Now put in the equation y=Ae^{Bt} and find A and B.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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