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(C) Given,$y^{2}=2 d(x+\sqrt{d})$  $(i)$Differentiating with respect to $x$,we get:$2y y_{1} = 2d \Rightarrow d = y y_{1}$Substituting $d = y y_{1}$ i

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degree $3$

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(C) Given,$y^{2}=2 d(x+\sqrt{d})$  $(i)$Differentiating with respect to $x$,we get:$2y y_{1} = 2d \Rightarrow d = y y_{1}$Substituting $d = y y_{1}$ into equation $(i)$:$y^{2} = 2(y y_{1})(x + \sqrt{y y_{1}})$Rearranging the terms to isolate the radical:$y^{2} - 2y y_{1} x = 2y y_{1} \sqrt{y y_{1}}$Squaring both sides:$(y^{2} - 2y y_{1} x)^{2} = (2y y_{1})^{2} (y y_{1})$$(y^{2} - 2y y_{1} x)^{2} = 4(y y_{1})^{3}$The highest derivative present is $y_{1}$ (order $1$),and the highest power of the highest derivative is $3$. Therefore,the degree of the differential equation is $3$.

The differential equation representing the family of curves $y^{2}=2 d(x+\sqrt{d})$,where $d$ is a parameter,is of

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