The given differential equation may be written as
` (dy)/(dx) - (x)/(( 1- x ^(2))) * y = (x ^(2))/((1- x ^(2)))" "`... (i)
This is of the form ` (dy)/(dx) + Py = Q`, where `P = (-x)/((1-x ^(2))) and Q = (x ^(2))/((1 - x ^(2)))`
Thus, the given differential equation is linear.
`IF= e ^(int Pdx) = e ^(int (-x)/((1- x ^(2))) dx ) = e ^((1)/(2) int (-2x)/((1- x ^(2))) dx ) = e ^((1)/(2) log ( 1 - x ^(2)))`
` " " = e ^(log sqrt (1 - x^(2))) = sqrt (1- x ^(2))`
So, its solution is given by
` y xx IF = int (Qxx IF ) dx + C`,
i.e., `y xx sqrt (1- x ^(2)) = int (x^(2))/(( 1- x ^(2))) xx sqrt (1 - x ^(2)) dx + C `
` " " = int (x^(2))/(sqrt (1- x ^(2))) dx + C `
` " " int ({ - ( 1 -x ^(2)) + 1})/( sqrt (1 - x ^(2))) dx + C `
` " " = - int sqrt (1 - x ^(2)) dx + int (1)/(sqrt (1 - x ^(2))) dx +C `
`= -{ (x sqrt (1- x ^(2)))/(2) + (1)/(2) sin ^(-1) x } + sin ^(-1) x + C `
` " " = (-x sqrt (1- x ^(2)))/( 2 ) + (1)/(2) sin ^(-1) x +C . `
` therefore y = (-x)/(2) + (sin ^(-1)x )/(2 sqrt (1- x ^(2))) + (C)/( sqrt (1- x ^(2)))" " ` ... (ii)
It is being given that when ` x = 0 `, then `y = 2 `
Putting ` x=0 and y =2 `in (ii), we get C = 2
Hence , `y = (-x)/(2) + (sin ^(-1)x)/(2sqrt( 1- x ^(2))) + (2)/(sqrt (1 - x ^(2)))` is the required solution.
