1. ## Separation of Variables

Use separation of variables to solve the IVP

How would you go about doing this? So far I've done this:

$\displaystyle dy/\sqrt(1-y^2) = dt/t$

$\displaystyle arcsin(y) = ln(t) +c$

I don't know if my work is correct, but would the solution just be $\displaystyle y=sin(ln(t))?$

2. Originally Posted by cdlegendary
Use separation of variables to solve the IVP
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How would you go about doing this? So far I've done this:

$\displaystyle dy/\sqrt(1-y^2) = dt/t$

$\displaystyle arcsin(y) = ln(t) +c$

I don't know if my work is correct, but would the solution just be $\displaystyle y=sin(ln(t))?$
Almost but not quite it does not satisfy the IC.

$\displaystyle y(1)=\sin(\ln(1))=\sin(0)=0$

You forgot the C. The solution should look like

$\displaystyle y=\sin(\ln(t)+c)$ Now use the IC to solve for c.

$\displaystyle 1=\sin(c) \implies c=\frac{\pi}{2}$

3. Originally Posted by TheEmptySet
Almost but not quite it does not satisfy the IC.

$\displaystyle y(1)=\sin(\ln(1))=\sin(0)=0$

You forgot the C. The solution should look like

$\displaystyle y=\sin(\ln(t)+c)$ Now use the IC to solve for c.

$\displaystyle 1=\sin(c) \implies c=\frac{\pi}{2}$
Ah I see, thanks. So essentially the $\displaystyle \pi/2$ makes it $\displaystyle cos(lnt)$