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`

Answer

`y*(1+x^2)=tan^(-1)x-pi/4`

Explanation

We have,


`(1+x^2)(dy)/(dx)+2x y=1/(1+x^2)`


`(dy)/(dx)+(2x)/(1+x^2) y=1/((1+x^2)^2)`


Comparing the equation with


`(dy)/(dx)+Py=Q`


Then, `P=(2x)/(1+x^2), quad Q=1/((1+x^2)^2)`


`I.F.=e^(int P dx) `


`=e^(int (2x)/(1+x^2) dx)=e^(log (1+x^2))=1+x^2`


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