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(D) Given the differential equation: $x \frac{d^2 y}{d x^2} = \left(1 + \left(\frac{d^2 y}{d x^2}\right)^2\right)^{-1/2}$.To find the degree,we must e

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$x^2 - 6x + 8 = 0$

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(D) Given the differential equation: $x \frac{d^2 y}{d x^2} = \left(1 + \left(\frac{d^2 y}{d x^2}\right)^2\right)^{-1/2}$.To find the degree,we must eliminate the negative exponent. Multiply both sides by $\left(1 + \left(\frac{d^2 y}{d x^2}\right)^2\right)^{1/2}$:$x \frac{d^2 y}{d x^2} \left(1 + \left(\frac{d^2 y}{d x^2}\right)^2\right)^{1/2} = 1$.Now,square both sides to remove the fractional exponent:$x^2 \left(\frac{d^2 y}{d x^2}\right)^2 \left(1 + \left(\frac{d^2 y}{d x^2}\right)^2\right) = 1$.Expanding this,we get $x^2 \left(\frac{d^2 y}{d x^2}\right)^2 + x^2 \left(\frac{d^2 y}{d x^2}\right)^4 = 1$.The highest order derivative present is $\frac{d^2 y}{d x^2}$,so the order $k = 2$.The highest power of the highest order derivative is $4$,so the degree $l = 4$.We need to find the quadratic equation whose roots are $k = 2$ and $l = 4$.The equation is $(x - 2)(x - 4) = 0$,which simplifies to $x^2 - 6x + 8 = 0$.

If the order and degree of the differential equation $x \frac{d^2 y}{d x^2} = \left(1 + \left(\frac{d^2 y}{d x^2}\right)^2\right)^{-1/2}$ are $k$ and $l$ respectively,then $k, l$ are the roots of

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