Mathematics - Differential Equation Question with Solution | TestHub

Let point$\mathrm{P}$is$(\mathrm{h},\mathrm{ }\mathrm{k})$
Now, equation of tangent at$\mathrm{P}$
$(y-k)=m_T (x- h)$...(i) ($m_T=$slope of tangent at$\mathrm{P}$)
For$Y_P$, put$x=0$in.....(i)
$Y_P \left(0, k- h{m}_T\right)$
Now, as given$P{Y}_P=1$
$\sqrt{h^2+h^2{m}_T^2}=1$
$\sqrt{h^2\left(1+m_T^2\right)}=1$
$1+m_T^2=\frac{1}{h^2}$
$m_T^2=\frac{1}{h^2}-1$
$m_T^2=\frac{1-h^2}{h^2}$
$m_T=\pm \frac{\sqrt{1-h^2}}{h}$
Put$m_T=\frac{dy}{dx}\mathrm{a}\mathrm{n}\mathrm{d}h=x$
$\frac{dy}{dx}=\pm \frac{\sqrt{1-x^2}}{x}$...(i)
$y=\pm \left(-\ln \left(\frac{1+\sqrt{1-x^2}}{x}\right)+\sqrt{1-x^2}\right)$
As the curve lies in$1^{\mathrm{s}\mathrm{t}}$quadrant y must be positive. Hence,
$y=\ln \left(\frac{1+\sqrt{1-x^2}}{x}\right)-\sqrt{1-x^2}$
Also from equation (i) only negative sign will give the correct equation of centre.
Hence,
$\frac{dy}{dx}=-\frac{\sqrt{1-x^2}}{x}$
$x{y}^{\prime }+\sqrt{1-x^2}=0$