# IVP problem

1. Apr 14, 2007

### ssb

1. The problem statement, all variables and given/known data
Please take a look at my work and help me figure out where I went wrong. Thanks!
Use separation of variables to solve the IVP
$$\[ty'=\sqrt{1-y^2},\quad y(1)=1,\quad t>0$$

2. Relevant equations

3. The attempt at a solution
Use separation of variables to solve the IVP
$$\[ty'=\sqrt{1-y^2},\quad y(1)=1,\quad t>0$$

$$dy/dt = \sqrt{1-y^2}/t$$

$$1/\sqrt{1-y^2} dy = 1/t dt$$

$$\int 1/\sqrt{1-y^2} dy = \int 1/t dt$$

$$arcsin(y) = ln (abs (t)) + C$$

$$C = 1$$

therefore the formula should be

$$y = sin(ln(abs(t))+1)$$

Thanks

Last edited: Apr 14, 2007
2. Apr 15, 2007

### Gib Z

I don't understand what you think is the problem.

3. Apr 15, 2007

### d_leet

I don't see a problem except that your value of C is incorrect, because with that choice y(1) is not equal to 1.

4. Apr 15, 2007

### HallsofIvy

Staff Emeritus
arcsin(1) is $\pi/2$, not 1.

I will also point out that this is of the form $y'= \sqrt{1- y^2}/t$ and the function is NOT Lipshchitz in y on any interval containing y= 1. Therefore, the solution is not unique. y= 1 for all t is another obvious solution and, in fact, there are an infinite number of solutions.