Is y = sqrt(x-1) a Solution to 2yy' = 1?

lap
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The question ask to determine whether y = φ(x) = sqrt ( x - 1 ) is a solution of the differential equation 2yy' = 1.
 
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As far as "determine whether y = φ(x) = sqrt ( x - 1 ) is a solution of the differential equation 2yy' = 1" is concerned, there is no reason to "rewrite the equation". If y= sqrt(x- 1)= (x-1)^(1/2) then y'= (1/2)(x- 1)^(-1/2) so that 2yy'= 2(x- 1)^(1/2)[(1/2)(x- 1)^(-1/2)= what?

As for the rest, I don't see how that has anything to do with the problem. Are you sure you haven't looked up the solution to the wrong question?
 
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Your wording is a bit confusing but if you're trying to determine if y(x) is a solution to your differential equation, all you have to do in plug y(x) into the equation and check if it holds.
 
lap has edited his question and changed it completely. My first response was to a different question.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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