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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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The differential equation of the family of curves $y^2=4a(x+a)$ is

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

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1 Answer

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Toolbox:
  • 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,$\large\frac{dv}{dx}=\frac{d^2y}{dx^2}$
Hence applying the product formula we get,
$y\large\frac{d^2y}{dx^2}+\large\frac{dy}{dx}.\frac{dy}{dx}$$=0 $
$=>y\large\frac{d^2y}{dx^2}+\bigg(\frac{dy}{dx}\bigg)^2=0 $
Hence the correct option is $C$
answered May 20, 2013 by meena.p
 

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