# The differential equation of the family of curves $y^2=4a(x+a)$ is

:$(A)\;y^2=4\frac{dy}{dx}\bigg(x+\frac{dy}{dx}\bigg) \quad (B)\;2y\frac{dy}{dx}+4a \quad(C)y\frac{d^2y}{dx^2}\;\bigg(\frac{dy}{dx}\bigg)^2=0 \quad (D)\;2x\frac{dy}{dx}+y\bigg(\frac{dy}{dx}\bigg)^2-y$

• If the given equation has $'n'$ arbitary constant, then its differential equation will be of $n^{th}$ order.
• $\large\frac{d}{dx}$$(uv)=\large\frac{dy}{dx}$$.v+\large\frac{dv}{dx}.$$u Given y^2=4a(x+a) Differentiating on both sides we get, 2y.\large\frac{dy}{dx}$$=4a=>y.\large\frac{dy}{dx}$$=2a Differentiating again w.r.t x, \large\frac{d}{dx}$$(uv)=\large\frac{d}{dx}$$(u).v+\large\frac{d}{dx}$$(v).u$
Let $u=y$ and on differentiating this we get $\large\frac{du}{dx}=\frac{dy}{dx}$
Let $v=\large\frac{dy}{dx},$ on differentiating we get,$\large\frac{dv}{dx}=\frac{d^2y}{dx^2}$