The differential equations, find a particular solution satisfying the given condition: `(1+x^2)(dy)/(dx)+2x y=1/(1+x^2); y=0`when `x = 1`

Explanation

`(1+x^(2))(dy)/(dx)+2xy=(1)/(1+x^(2))`
`implies (dy)/(dx)+(2x)/(1+x^(2))y=(1)/((1+x^(2))^(2))`
Here, `P=(2x)/(1+x^(2))` and `Q=(1)/((1+x^(2))^(2))`
`I.F.=e^(intPdx)=e^(int(2x)/(1+x^(2))dx)`
`:. =e^(log(1+x^(2))=(1+x^(2))`
and general solution :
`y*(1+x^(2))=int(1)/((1+x^(2))^(2))(1+x^(2))dx+c`
`=int(1)/(1+x^(2))dx+c`
`y(1+x^(2))=tan^(-1)x+c`.........`(1)`
Given, `y=0` at `x=1`
`0=tan^(-1)1+c`
`impliesc=-(pi)/(4)`
Therefore, from equation `(1)`, particular solution is
`y(1+x^(2))=tan^(-1)x-(pi)/(4)`