Hello.
I'm stuck at this differential equation:
$\displaystyle x\cdot y'+y^{2}=1,$ $\displaystyle y(1)=0$
Not even sure where to start, because everything I tried failed.
Appreciate any help.
Do you know if your answer is correct?
I was thinking
$\displaystyle \frac{1}{2}\ln(y+1)-\frac{1}{2}\ln(y-1) = \ln x +C $
$\displaystyle \frac{1}{2}\ln\frac{(y+1)}{(y-1)} = \ln x +\ln C $
$\displaystyle \ln\sqrt{\frac{(y+1)}{(y-1)}} = \ln x +\ln C $
$\displaystyle \ln\sqrt{\frac{(y+1)}{(y-1)}} = \ln (Cx) $
$\displaystyle \sqrt{\frac{(y+1)}{(y-1)}} = Cx $
$\displaystyle \frac{(y+1)}{(y-1)} = Cx^2 $
$\displaystyle \frac{2}{(y-1)}+1 = Cx^2 $
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$\displaystyle y = \frac{2}{Cx^2-1}+1$
Then use the initial condition $\displaystyle y(1)=0$ to find $\displaystyle C$