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  • The differential equation of family of curves y^{2}=4 a(x+a) is (a) y^{2}=4 \frac{d y}{d x} x+\frac{d y}{d x}

The differential equation of family of curves y^{2}=4 a(x+a) is
(a) y^{2}=4 \frac{d y}{d x}\left(x+\frac{d y}{d x}\right)
(b) 2 y \frac{d y}{d x}=4 a
(c) \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}=0
(d) 2 x \frac{d y}{d x}+y\left(\frac{d y}{d x}\right)^{2}-y=0

Answers (1)

Ans: - The answer is the option (d) 2 x \frac{d y}{d x}+y\left(\frac{d y}{d x}\right)^{2}-y=0$

Explanation: -

y^{2}=4 a(x+a)$

On differentiating both sides w.r.t. x, we get
$$ 2 y \frac{d y}{d x}=4 a $$
\Rightarrow \quad \frac{1}{2} y \frac{d y}{d x}=a$
On putting the value of a in Eq. (i), we get
$$ y^{2}=2 y \frac{d y}{d x}\left(x+\frac{1}{2} y \frac{d y}{d x}\right) $$
\Rightarrow \quad y^{2}=2 x y \frac{d y}{d x}+y^{2}\left(\frac{d y}{d x}\right)^{2}$
$$ \Rightarrow \quad 2 x \frac{d y}{d x}+y\left(\frac{d y}{d x}\right)^{2}-y=0 $$

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