A differential equation of the form \(y=p x+f(p)\), where \(p=\frac{d y}{d x}\) is
known as Clairaut's equation. We now find the solution of the above equation.
The given equation is
$$
y=p x+f(p)
$$
It is solvable for \(y\). Differentiating with respect to \(x\), we get
$$
\frac{d y}{d x}=p+x \frac{d p}{d x}+f^{\prime}(p) \cdot \frac{d p}{d x}
$$
\(\Rightarrow\)
$$
p=p+x \frac{d p}{d x}+f^{\prime}(p) \cdot \frac{d p}{d x}
$$
\(\Rightarrow \quad x \frac{d p}{d x}+f^{\prime}(p) \cdot \frac{d p}{d x}=0\)
Factorizing \(\frac{d p}{d x}+\left[x+f^{\prime}(p)\right]=0\)
Cancelling the factor \(x+f^{\prime}(p)\) which does not involve \(\frac{d p}{d
x}\), we have \(\frac{d p}{d x}=0\)
Integrating,
$$
p=c
$$
Eliminating \(p\) between (1) and (2), the required solution of \((1)\) is
$$
y=c x+f(c)
$$
The solution of the equation \(\left(\frac{d y}{d
x}\right)^{2}\left(x^{2}-a^{2}\right)-2\left(\frac{d y}{d x}\right) x
y+y^{2}-b^{2}=0\) is
(A) \(y=c x+\sqrt{a^{2} c^{2}+b^{2}}\)
(B) \(y=c x+\sqrt{b^{2} c^{2}+a^{2}}\)
(C) \(y=c x-\sqrt{a^{2} c^{2}+b^{2}}\)
(D) \(y=c x-\sqrt{b^{2} c^{2}+a^{2}}\)