Q3.    Form the differential equation representing the family of curves given by (x-a)^2 + 2y^2 = a^2 , where a is an arbitrary constant.

Answers (1)

Given equation is
(x-a)^2 + 2y^2 = a^2                   
we can rewrite it as
2y^2 = a^2-(x-a)^2                   -(i)
Differentiate both the sides w.r.t x
\frac{d\left ( 2y^2 \right )}{dx}=\frac{d(a^2-(x-a)^2)}{dx}
4yy^{'}=4y\frac{dy}{dx}=-2(x-a)\\ \\                 
(x-a)= -2yy'\Rightarrow a = x+2yy'                          -(ii)
Put value from equation (ii)  in (i)
 (-2yy')^2+2y^2= (x+2yy')^2\\ 4y^2(y')^2+2y^2= x^2+4y^2(y')^2+4xyy'\\ y' = \frac{2y^2-x^2}{4xy}
Therefore, the required differential equation is y' = \frac{2y^2-x^2}{4xy}                   
 

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