Solution:
assuming (p, 0) be the centre of the circle.
Therefore, it touches the y – axis at origin, its radius is p.
Since, the equation of the circle with centre (p, 0) and radius (p) is
\[\begin{array}{*{35}{l}}
\Rightarrow {{\left( x\text{ }-\text{ }p \right)}^{2}}~+\text{ }{{y}^{2}}~=\text{ }{{p}^{2}} \\
\Rightarrow {{x}^{2}}~+\text{ }{{p}^{2}}~-\text{ }2xp\text{ }+\text{ }{{y}^{2}}~=\text{ }{{p}^{2}} \\
\end{array}\]
Transposing p2 and – 2xp to RHS then it becomes – p2 and 2xp
\[\begin{array}{*{35}{l}}
\Rightarrow {{x}^{2}}~+\text{ }{{y}^{2}}~=\text{ }{{p}^{2}}~-\text{ }{{p}^{2}}~+\text{ }2px \\
\Rightarrow ~{{x}^{2}}~+\text{ }{{y}^{2}}~=\text{ }2px\text{ }\ldots \text{ }\left[ equation\text{ }\left( i \right) \right] \\
\end{array}\]
differentiating equation (i) both sides,
\[\begin{array}{*{35}{l}}
\Rightarrow 2x\text{ }+\text{ }2yy\text{ }=\text{ }2p \\
\Rightarrow ~x\text{ }+\text{ }yy\text{ }=\text{ }p \\
\end{array}\]
substituting the value of ‘p’ in the equation, we get,
\[\begin{array}{*{35}{l}}
\Rightarrow {{x}^{2}}~+\text{ }{{y}^{2}}~=\text{ }2\left( x\text{ }+\text{ }yy \right)x \\
\Rightarrow ~2xyy\text{ }+\text{ }{{x}^{2}}~=\text{ }{{y}^{2}} \\
\end{array}\]
Hence, 2xyy’ + x2 = y2 is the required differential equation.