The difference between the degree and the ord | vedclass

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Question

(D) Given the family of curves: $y^{2}=a\left(x+\frac{\sqrt{a}}{2}\right) = ax + \frac{a^{3/2}}{2} \quad ...(1)$Differentiating with respect to $x$:$2

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$2$

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(D) Given the family of curves: $y^{2}=a\left(x+\frac{\sqrt{a}}{2}\right) = ax + \frac{a^{3/2}}{2} \quad ...(1)$Differentiating with respect to $x$:$2yy' = a$Substitute $a = 2yy'$ into equation $(1)$:$y^{2} = (2yy')x + \frac{(2yy')^{3/2}}{2}$Rearrange the terms:$y^{2} - 2xyy' = \frac{(2yy')^{3/2}}{2}$Square both sides to eliminate the fractional exponent:$(y^{2} - 2xyy')^{2} = \frac{(2yy')^{3}}{4}$$(y^{2} - 2xyy')^{2} = 2y^{3}(y')^{3}$The highest order derivative present is $y'$,so the order is $1$.The highest power of the highest order derivative is $3$,so the degree is $3$.The difference between the degree and the order is $3 - 1 = 2$.

The difference between the degree and the order of the differential equation that represents the family of curves given by $y^{2}=a\left(x+\frac{\sqrt{a}}{2}\right)$,where $a>0$,is:

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