Solving a second order ODE using reduction of order

Bonnie
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Homework Statement


Hi there, I have an assignment which involves using reduction of order to solve for a second solution to an ode (the one attached). However this is a method I am new to, and though I have tried several times, I'm somehow getting something wrong because the LHS and RHS are not matching up, that is, when I substitue in the solution I have found, the RHS does not equal zero as it should.

Homework Equations

The Attempt at a Solution


I have attached my working (Sides 1 and 2), if anyone could point out what I'm doing wrong it would be greatly appreciated, this is driving me nuts!
 

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According to wolfram your general solution is correct.

You do a blunt mistake in the check, you put ##y'=-\frac 1 2 t^{-\frac 1 2}##, the minus in front is not needed. It is clear that ##y'=\frac{1}{2} t^{-\frac 1 2}## for ##y=t^{\frac 1 2}##
 
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Delta² said:
According to wolfram your general solution is correct.

You do a blunt mistake in the check, you put ##y'=-\frac 1 2 t^{-\frac 1 2}##, the minus in front is not needed. It is clear that ##y'=\frac{1}{2} t^{-\frac 1 2}## for ##y=t^{\frac 1 2}##
Oh, that was dumb. Thank you!
 
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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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