Mathematics - Differential Equation Question with Solution | TestHub

Equation of normal at the point $p(x,y)$ is $Y-y=-\frac{dx}{dy}(X-x)\left(\mathrm{let}m=\frac{dx}{dy}\right)$

Let, $\mathrm{m}=\frac{\mathrm{dx}}{\mathrm{dy}}\Rightarrow X+\mathrm{mY}-(\mathrm{x}+\mathrm{my})=0$

Distance of perpendicular from the origin to line (i) is $\frac{|x+my|}{\sqrt{1+m^2}}=|y|$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{{\mathrm{y}}^2-{\mathrm{x}}^2}{2\mathrm{xy}}$

This is homogeneous equation

Let, $\mathrm{y}=\mathrm{zx}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{z}+\mathrm{z}\frac{\mathrm{dz}}{\mathrm{dx}}$

$\Rightarrow \frac{2\mathrm{z}}{1+{\mathrm{z}}^2}\mathrm{dz}=-\frac{\mathrm{dx}}{\mathrm{x}}$

Integrating

$\int \frac{2z}{1+z^2}dz=-\int \frac{dx}{x}$

$\Rightarrow \log \left(1+{\mathrm{z}}^2\right)=-\log \mathrm{x}+\mathrm{c}$

$\Rightarrow \left({\mathrm{x}}^2+{\mathrm{y}}^2\right)=\mathrm{x}.{\mathrm{e}}^{\mathrm{c}}$

This curve passes through $(1,1)$

$\Rightarrow 1+1=1.{\mathrm{e}}^{\mathrm{c}}$

$e^c=2$

The required equation of the curve is

$\Rightarrow x^2+y^2=2x$