Final round: Integrating factors

ssb
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Homework Statement


Final round I promise!

Is there some sort of trick that can be applied to the following equation so that it is easier to process?
\frac{dy}{dt}=\frac{1}{t+y},\:y(-1)=0

The Attempt at a Solution



Somebody told me that the equation can be made easier by reinterpreting it with y as the independent variable and t as the dependent variable. I still have a vacant look about my face now as I did back then.

I know that \frac{dy}{dt}=\frac{1}{\frac{dt}{dy}} but when I try to apply it to my initial equation, I get the same thing I started with.
 
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'Somebody' is right. It turns the equation into 1/t'=1/(t+y). Or t'=t+y. That does look a bit simpler, right?
 
Dick said:
'Somebody' is right. It turns the equation into 1/t'=1/(t+y). Or t'=t+y. That does look a bit simpler, right?

Then you just solve it the normal way as if t' were y'? I don't mean rename everything but just work the problem through solving for f(y) instead of y(t)?

Just fyi I love you.
 
ssb said:
Then you just solve it the normal way as if t' were y'? I don't mean rename everything but just work the problem through solving for f(y) instead of y(t)?

Just fyi I love you.

Exactly...
 
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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