Question

In problems 23-28 Find an explicit solution of the given initial-value problem.

$\sqrt{{\text{1 -y}}^{\text{2}}}\text{dx -}\sqrt{{\text{1 -x}}^{\text{2}}}\text{dy = 0,}\text{y(0) =}\frac{\sqrt{\text{3}}}{\text{2}}$

Short answer

$y=\frac{1}{2}x+\frac{\sqrt{3}}{2}\sqrt{1-x^2}$

Step-by-step solution

Definition of separable equation

A first-order differential equation of the form $\frac{dy}{dx}=g\left(x\right)h\left(y\right)$is said to be separable or to have separable variables.

Separate the variables

Rearrange and separate the variable

$$\begin{gathered} \sqrt{1-x^2}dy=\sqrt{1-y^2}dx \\ \frac{dy}{\sqrt{1-y^2}}=\frac{dx}{\sqrt{1-x^2}} \end{gathered}$$

Integrate both sides

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

Integration

Do the integration

${\sin }^{-1}y={\sin }^{-1}x+C$

Substituting initial condition

$y\left(0\right)=\frac{\sqrt{3}}{2}$

Therefore, the solution is as follows

$$\begin{gathered} {\sin }^{-1}\left(\frac{\sqrt{3}}{2}\right)={\sin }^{-1}\left(0\right)+C \\ \frac{\pi }{3}=C \end{gathered}$$

Therefore, the solution is as follows

${\sin }^{-1}y={\sin }^{-1}x+\frac{\pi }{3}$

Simplification

Simplifying the solution using the formula:

$y=\sin \left({\sin }^{-1}x+\frac{\pi }{3}\right)$

$\sin (a+b)=\sin a\cos b+\cos a\sin b$and $\cos \left({\sin }^{-1}x\right)=\sqrt{1-x^2}$

Then

$$\begin{gathered} y=\sin \left({\sin }^{-1}x\right)\cos \left(\frac{\pi }{3}\right)+\cos \left({\sin }^{1}x\right)\sin \left(\frac{\pi }{3}\right) \\ y=\frac{1}{2}x+\frac{\sqrt{3}}{2}\sqrt{1-x^2} \end{gathered}$$

Hence, the solution is$\text{y =}\frac{\text{1}}{\text{2}}\text{x +}\frac{\sqrt{\text{3}}}{\text{2}}\sqrt{{\text{1 -x}}^{\text{2}}}$