Question

Use separation of variables to obtain
solutions to the DEs and IVPs. Solve for y when possible.

Answer 1

\[ty'=\sqrt{1-y ^{2}}\]

Answer 2

Did you have any ideas on this question?

Answer 3

I gotta go, so I'll leave with this:
The whole right side is an expression in y. So we could move that over with the y'. The t multipled onto y' is just the t term, so it can be moved to the right side. That gives us the form: F(y) dy = G(x) dx, where we can integrate both sides.

The left side you might use trig substitution or recognize that it is the derivative of arcsine. The right side is the natural log integral.

Answer 4

\[t\frac{dy}{dt}=\sqrt{1-y^2}~~\iff~~\frac{dy}{\sqrt{1-y^2}}=\frac{dt}{t}\]

Answer 5

when you do the integration how do the results look like?

Answer 6

@sithsandgiggles

Answer 7

For the right side, let \(y=\sin u\), so \(dy=\cos u~du\).
\[\int\frac{dy}{\sqrt{1-y^2}}=\int\frac{\cos u}{\sqrt{1-\sin^2u}}~du=\int\frac{\cos u}{\sqrt{\cos^2u}}~du=\int du\]
Use the fact that \(y=\sin u~~\iff~~u=\sin^{-1}y\) when you integrate.
\[\int du=\int\frac{dt}{t}\]
I'll stop right here.

Answer 8

is that the answer?

Answer 9

There is further simplification to be made, but it is close. Carry out those integrals and back-substitute your u = sin^-1 y. Then you can solve for y.

Answer 10

im confused?

Answer 11

@accessdenied

Answer 12

Well, your end goal is to get to your original function y. Where we left off:
\( \displaystyle \int du = \int \frac{dt}t \quad u = \sin ^{-1} y \)
These integrals will simplify further; we cannot leave it here! The left side is simply equal to u, and the right side is the natural logarithm function. Once we've evaluated integrals, we plug the value for u back in and then solve for y.

Answer 13

so is it sin^-1(y)=ln(t)

Answer 14

Almost! Just add in the constant of integration:
sin^-1 (y) = ln t + c
And since the directions ask to solve for y when possible, we can also take the sine of both sides to cancel the inverse sine:
y = sin(ln t + c)