The order and degree of the differential equa | vedclass

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(D) Given the differential equation: $\left(\frac{d^3 y}{d x^3}\right)^{5/4} = \left(\frac{d^2 y}{d x^2}\right)^{4/3}$.To eliminate the fractional exp

Vedclass · Accepted Answer

$3, 15$

Vedclass · Answer

(D) Given the differential equation: $\left(\frac{d^3 y}{d x^3}\right)^{5/4} = \left(\frac{d^2 y}{d x^2}\right)^{4/3}$.To eliminate the fractional exponents,raise both sides to the power of $12$ (the least common multiple of $4$ and $3$):$\left(\left(\frac{d^3 y}{d x^3}\right)^{5/4}\right)^{12} = \left(\left(\frac{d^2 y}{d x^2}\right)^{4/3}\right)^{12}$.This simplifies to: $\left(\frac{d^3 y}{d x^3}\right)^{15} = \left(\frac{d^2 y}{d x^2}\right)^{16}$.The highest order derivative present is $\frac{d^3 y}{d x^3}$,so the order is $3$.The degree is the power of the highest order derivative after the equation is made free from radicals and fractions,which is $15$.Thus,the order is $3$ and the degree is $15$.

The order and degree of the differential equation $\sqrt[4]{\left(\frac{d^3 y}{d x^3}\right)^5} = \sqrt[3]{\left(\frac{d^2 y}{d x^2}\right)^4}$ are . . . . . . and . . . . . . respectively.

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