Question

The solution curve of the differential equation $(1+x^2)\frac{dy}{dx}+2xy=\frac{1}{(1+x^2)}$ passing though the point (1, 0) is:

Answer 1

Given solution of curve $(1+x^2)\frac{dy}{dx}+2xy=\frac{1}{1+x^2}$ passes through (1,0) then

$(1+x^2)\frac{dy}{dx}+2xy=\frac{1}{1+x^2}$ _________ (1)

$\frac{dy}{dx}+\frac{2x}{1+x^2}.y=\frac{1}{(1+x^2{)}^2}$

Using integrating factor concept

$I.F=e^{\int \frac{2x}{1+x^2}dx}$ $e^{log(1+x^2)}=1+x^2$

$[\int \frac{2x}{1+x^2}dx=log(1+x^2\left)\right]$

nw solution of differential equation (1) is

$y\times (I.F)=\int \frac{1}{(1+x^2{)}^2}\times (I.F).dx$

$y(1+x^2)=\int \frac{1}{(1+x^2{)}^2}\times (1+x^2)dx$

$y(1+x^2)=\int \frac{1}{1+x^2}dx$

$y(1+x^2)=ta{n}^{-1}x+c$ $[\int \frac{1}{1+x^2}dx=ta{n}^{-1}x]$

Now (1,0) is a point in curve then

$0(1+1)=ta{n}^{-1}\left(1\right)+c$

$c=-\frac{\pi }{4}$

now equation of curve is

$y(1+x^2)=ta{n}^{-1}x-\frac{\pi }{4}$