Answer:
The particular solution is [tex]y=\sin (\ln|x|)[/tex] .
Step-by-step explanation:
The given differential equation is
[tex]xy'=\sqrt {1-y^2}[/tex]
It can be written as
[tex]x\frac{dy}{dx}=\sqrt {1-y^2}[/tex]
Use variable separable method to solve the above equation.
[tex]\frac{dy}{\sqrt {1-y^2}}=\frac{1}{x}dx[/tex]
Integrate both sides.
[tex]\int \frac{dy}{\sqrt {1-y^2}}=\int \frac{1}{x}dx[/tex]
[tex]\sin^{-1} y=\ln|x|+C[/tex] .... (1)
It is given that y(1)=0. It means y=0 at x=1.
[tex]\sin (0)=\ln|1|+C[/tex]
[tex]0=0+C[/tex]
[tex]0=C[/tex]
The value of constant is 0.
Substitute C=0 in equation (1) to find The required equation.
[tex]\sin^{-1} y=\ln|x|+0[/tex]
Taking sin both sides.
[tex]y=\sin (\ln|x|)[/tex]
Therefore the particular solution is [tex]y=\sin (\ln|x|)[/tex] .
