Question

The function $y=sinx$ is an explicit solution of the first-order differential equation$\frac{dy}{dx}=\sqrt{1-y^2}$ . Find an interval $I$of definition. [Hint: $I$is not the interval $(-\infty ,\infty )$.]

Short answer

$y\text{'}$ is defined along the interval $(-1,1]$ because the solution converges about $1$ and diverges about $-1$.So,$I:(-1,1]$ .

Step-by-step solution

Step 1:Definition of an explicit solution.

In explicit solution unknown functions (dependable variables) is indicated (expressed) in terms of undependable variable and constants. This is known as “explicit solution”.

Use of the method of separation of variables.

Given that,$\frac{dy}{dx}=\sqrt{1-y^2}$

Solve for the$y$by use of the method of separation of variables.

$\frac{dy}{\sqrt{1-y^2}}=dx$

Take integral on both sides,

$\int \frac{1}{\sqrt{1-y^2}}dy=\int dx$

We get,

$$\begin{gathered} {\sin }^{-1}y=x+C \\ y\left(x\right)=\sin (x+C) \end{gathered}$$

Therefore,$y\text{'}$is defined along the interval$(-1,1]$because the solution converges about$1$and diverges about$-1$.So,$I:(-1,1]$