Solution of the differential equation $(1+y^2-x^2)y'=\frac{1}{x}, y(1)=1$ Solution of the differential equation $(1+y^2-x^2)y'=\frac{1}{x}, y(1)=1$ is the differential equation have solution ? and is this solution is exist?
I am trying to prove by using picard's theorem 
here given ODE is $(1+y^2-x^2)y'=\frac{1}{x}$
then $f(x,y)=\frac{1}{(1+y^2-x^2)x}\Rightarrow |f(x,y)|=|\frac{1}{(1+y^2-x^2)x}|$  from here how to processed
 A: $(1+y^2-x^2)y'=\dfrac{1}{x}$ with $y(1)=1$
$(1+y^2-x^2)\dfrac{dy}{dx}=\dfrac{1}{x}$ with $y(1)=1$
$\dfrac{1}{x}\dfrac{dx}{dy}=1+y^2-x^2$ with $x(1)=1$
Let $u=\dfrac{1}{x^2}$ ,
Then $\dfrac{du}{dy}=-\dfrac{2}{x^3}\dfrac{dx}{dy}$
$\therefore-\dfrac{x^2}{2}\dfrac{du}{dy}=1+y^2-x^2$ with $u(1)=1$
$\dfrac{du}{dy}=2-\dfrac{2(y^2+1)}{x^2}$ with $u(1)=1$
$\dfrac{du}{dy}=2-2(y^2+1)u$ with $u(1)=1$
$\dfrac{du}{dy}+2(y^2+1)u=2$ with $u(1)=1$
I.F. $=e^{\int2(y^2+1)~dy}=e^{\frac{2y^3}{3}+2y}$
$\therefore\dfrac{d\left(ue^{\frac{2y^3}{3}+2y}\right)}{dy}=2e^{\frac{2y^3}{3}+2y}$ with $u(1)=1$
$u=2e^{-\frac{2y^3}{3}-2y}\int e^{\frac{2y^3}{3}+2y}~dy+Ce^{-\frac{2y^3}{3}-2y}$ with $u(1)=1$
$Ce^{-\frac{2}{3}-2}=1$
$C=e^\frac{8}{3}$
$\therefore u=2e^{-\frac{2y^3}{3}-2y}\int_1^ye^{\frac{2y^3}{3}+2y}~dy+e^{\frac{8}{3}-2y-\frac{2y^3}{3}}$
$\dfrac{1}{x^2}=2e^{-\frac{2y^3}{3}-2y}\int_1^ye^{\frac{2y^3}{3}+2y}~dy+e^{\frac{8}{3}-2y-\frac{2y^3}{3}}$
$x^2=\dfrac{1}{2e^{-\frac{2y^3}{3}-2y}\int_1^ye^{\frac{2y^3}{3}+2y}~dy+e^{\frac{8}{3}-2y-\frac{2y^3}{3}}}$
