Question

Use separation of variables to obtain solutions to the DEs and IVPs. Solve for y when possible. y(0)=-1

Answer 1

\[y'=\frac{ -1+y ^{2} }{ 1+t^2 } \]

Answer 2

\[\begin{align*}
\frac{dy}{y^2-1}&=\frac{dt}{1+t^2}
\end{align*}\]
Left side: sub \(y=\sec r\), so \(dy=\sec r\tan r~dr\).
Right side: sub \(t=\tan s\), so \(dt=\sec^2s~ds\).

Integrating both sides, you have
\[\begin{align*}
\int\frac{\sec r\tan r}{\sec^2r-1}~dr&=\int\frac{\sec^2s}{1+\tan^2s}~ds
\end{align*}\]

Answer 3

\[\begin{align*}
\int\frac{\sec r\tan r}{\tan^2r}~dr&=\int\frac{\sec^2s}{\sec^2s}~ds\\\\
\int\frac{\sec r}{\tan r}~dr&=\int ds\\\\
\int\frac{\frac{1}{\cos r}}{\frac{\sin r}{\cos r}}~dr&=\int ds\\\\
\int\frac{1}{\sin r}~dr&=\int ds\\\\
\int\csc r~dr&=\int ds\\
&\vdots
\end{align*}\]

Answer 4

is that the answer?

Answer 5

@SithsAndGiggles

Answer 6

No, it's not the answer. It's MOST of the work needed to get the answer. There's a reason I left it open-ended like I did.

Answer 7

what do i do next, i don't understand

Answer 8

@sithsandgiggles