Question

Xy′ = √(1 − y2 ), y(1) = 0

Answer 1

Answer:

The particular solution is $y=\sin (\ln|x|)$ .

Step-by-step explanation:

The given differential equation is

$xy'=\sqrt {1-y^2}$

It can be written as

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

Use variable separable method to solve the above equation.

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

Integrate both sides.

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

$\sin^{-1} y=\ln|x|+C$            .... (1)

It is given that y(1)=0. It means y=0 at x=1.

$\sin (0)=\ln|1|+C$

$0=0+C$

$0=C$

The value of constant is 0.

Substitute C=0 in equation (1) to find The required equation.

$\sin^{-1} y=\ln|x|+0$

Taking sin both sides.

$y=\sin (\ln|x|)$

Therefore the particular solution is $y=\sin (\ln|x|)$ .