Question

Question: Find a function whose square plus the square of its derivative is 1.

Short answer

Answer:

The functions are.$y=\sin (x+c) and y=\sin (c-x)$.

Step-by-step solution

Definition

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

Find function

Suppose the function$y\left(x\right)$is such that$y^2+{\left(\frac{dy}{dx}\right)}^2=1$

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

Separate the variables and in the differential equation.

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

Integrate both sides.

$$\begin{gathered} \int \frac{dy}{\sqrt{1-y^2}}=\pm \int dx \\ {\sin }^{-1}y=\pm x+c \\ y=\sin (\pm x+c) \end{gathered}$$

Hence, there are two functions whose square plus the square of its derivative is 1.