Find the particular solution of the differential equation ` (sqrt (1-y^(2)) dx = (sin ^(-1)y - x ) dy `, it being given that when `y=0`, then ` x =0`.

Explanation

The given differential equation may be written as
` (dx)/(dy) = (sin ^(-1) y- x )/( sqrt (1- y^(2)))`
`rArr (dx)/(dy ) + (1 )/(sqrt (1 -y ^(2))) *x = (sin ^(-1) y)/(sqrt(1- y^(2)))" " `... (i)
This is of the form ` (dx)/(dy + Px = Q`, where ` P = (1)/(sqrt (1- y^(2))) and Q = (sin ^(-1) y )/( sqrt (1 - y^(2)))`
Thus, the given equation is linear.
`If = e ^(int Pdy) = e^(int (1)/(sqrt (1+ y^(2)))dy) = e ^(sin ^(-1)y)`
So, the solution of the given differential equation is given by
`x xx IF = int (Q xx IF )dy +C`,
i.e., `x xx e ^(sin ^(-1)y) = int {(sin ^(-1)y)/(sqrt (1-y^(2))) * e ^( (sin ^(-1) y )) } dy + C `
` " " = int underset ("I") t underset ("II") e ^(t) dt +C, ` where ` sin ^(-1)y =t and (1)/(sqrt (1-y^(2)))dy =dt `
` " " = t e ^(t) - int 1*e ^(t) dt +C `
` " " = t e ^(t) - int e ^(t) dt +C = ( te ^(t) - e ^(t)) + C `
` = ( t- 1 ) e ^(t) +C `
` = (sin ^(-1)y - 1 ) e ^(sin ^(-1) y )+C " " [because t = sin ^(-1 ) y ]`
`therefore x = (sin^(-1) y - 1) + Ce^(-sin ^(-1) y )" "`...(ii)
Putting y =0 and x =0 in (ii), we get C = 1
Hence, ` x= (sin ^(-1) y - 1 ) + e ^(-sin^(-1)y) ` is the required solution.