# Separation of Variables

• Jun 6th 2010, 01:32 PM
cdlegendary
Separation of Variables
Use separation of variables to solve the IVP
http://homework.math.ucsb.edu/webwor...e4c4b70131.png
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 $\displaystyle y=sin(ln(t))?$
• Jun 6th 2010, 01:45 PM
TheEmptySet
Quote:

Originally Posted by cdlegendary
Use separation of variables to solve the IVP
http://homework.math.ucsb.edu/webwor...e4c4b70131.png,http://homework.math.ucsb.edu/webwor...e4c4b70131.png,http://homework.math.ucsb.edu/webwor...e4c4b70131.png

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.

$\displaystyle 1=\sin(c) \implies c=\frac{\pi}{2}$
• Jun 6th 2010, 01:51 PM
cdlegendary
Quote:

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 $\displaystyle cos(lnt)$