Solving IVP: \[ty'=\sqrt{1-y^2},\quad y(1)=1,\quad t>0

[ssb]

Homework Statement

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$$

Homework Equations

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

 

[Gib Z]

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

 

[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.

 

[HallsofIvy]

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.