Finding Solutions for Second Order ODE with Initial Condition y(0)=6

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Homework Statement



How many functions y(t) satisfy both y''+t^2*y=0 and y(0)=6?

2. The attempt at a solution

As this is a second order differential equation, two initial conditions (for y and y') would be needed to obtain a unique solution (cf. existence and uniqueness theorem). So the answer is 'infinitely many functions' as we are given only y(0)=6. Have I understood it correctly?
 
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As long as all those different y'(0) lead to valid solutions, that works.
 
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mfb said:
As long as all those different y'(0) lead to valid solutions, that works.

Oh, OK. Many thanks. :)
 

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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