The order and degree of the differential equa | vedclass

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Question

(D) Given the differential equation: $\frac{d^2 y}{d x^2}+y+\left(\frac{d y}{d x}-\frac{d^3 y}{d x^3}\right)^{3 / 2}=0$. Rearranging the terms, we get

Vedclass · Accepted Answer

$3, 3$

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(D) Given the differential equation: $\frac{d^2 y}{d x^2}+y+\left(\frac{d y}{d x}-\frac{d^3 y}{d x^3}\right)^{3 / 2}=0$. Rearranging the terms, we get: $\left(\frac{d y}{d x}-\frac{d^3 y}{d x^3}\right)^{3 / 2} = -\left(\frac{d^2 y}{d x^2}+y\right)$. Squaring both sides to eliminate the fractional exponent: $\left(\frac{d y}{d x}-\frac{d^3 y}{d x^3}\right)^3 = \left(\frac{d^2 y}{d x^2}+y\right)^2$. The highest order derivative present in the equation is $\frac{d^3 y}{d x^3}$, so the order is $3$. The power to which the highest order derivative is raised after the equation is made free from radicals and fractions is $3$. Therefore, the order is $3$ and the degree is $3$.

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

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