The order and degree of the differential equa | vedclass

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(B) The given differential equation is $\left(\frac{d^3 y}{d x^3}\right)^{\frac{1}{2}} = 2\left(\frac{d y}{d x}\right)^{\frac{1}{4}} - x y$. To find t

Vedclass · Accepted Answer

$3$ and $2$

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(B) The given differential equation is $\left(\frac{d^3 y}{d x^3}\right)^{\frac{1}{2}} = 2\left(\frac{d y}{d x}\right)^{\frac{1}{4}} - x y$. To find the degree,we must eliminate the fractional powers of the derivatives. The powers are $\frac{1}{2}$ and $\frac{1}{4}$. The least common multiple of the denominators $2$ and $4$ is $4$. Raising both sides to the power of $4$,we get: $\left(\left(\frac{d^3 y}{d x^3}\right)^{\frac{1}{2}}\right)^4 = \left(2\left(\frac{d y}{d x}\right)^{\frac{1}{4}} - x y\right)^4$ $\left(\frac{d^3 y}{d x^3}\right)^2 = \left(2\left(\frac{d y}{d x}\right)^{\frac{1}{4}} - x y\right)^4$. Expanding the right side using the binomial theorem,the highest derivative present is $\frac{d^3 y}{d x^3}$,so the order is $3$. The highest power of the highest order derivative $\frac{d^3 y}{d x^3}$ after clearing the radicals is $2$. Therefore,the order is $3$ and the degree is $2$.

The order and degree of the differential equation $\left(\frac{d^3 y}{d x^3}\right)^{\frac{1}{2}}-2\left(\frac{d y}{d x}\right)^{\frac{1}{4}}+x y=0$ are respectively

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