Find the particular solution of the differential equation(`tan^(-1)y-x)dy=(1+y^2)dx ,` given that when `x=0,\ y=0.`

Explanation

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