Separation of Variables

**cdlegendary** · June 6th 2010, 01:32 PM

Use separation of variables to solve the IVP

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

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

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

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

**TheEmptySet** · June 6th 2010, 01:45 PM

Originally Posted by cdlegendary

Use separation of variables to solve the IVP

,

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

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

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

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

Almost but not quite it does not satisfy the IC.

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

You forgot the C. The solution should look like

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

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

**cdlegendary** · June 6th 2010, 01:51 PM

Originally Posted by TheEmptySet

Almost but not quite it does not satisfy the IC.

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

You forgot the C. The solution should look like

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

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

Ah I see, thanks. So essentially the $\pi/2$ makes it $cos(lnt)$

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