The order and degree of the differential equa | vedclass

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Question

(A) Given the differential equation: $\frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^{1/3} + x^{1/4} = 0$. To find the degree,we must eliminate the

Vedclass · Accepted Answer

$2, 3$

Vedclass · Answer

(A) Given the differential equation: $\frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^{1/3} + x^{1/4} = 0$. To find the degree,we must eliminate the fractional exponents. Rewrite the equation as: $\frac{d^2y}{dx^2} + x^{1/4} = -\left( \frac{dy}{dx} \right)^{1/3}$. Cube both sides to remove the exponent $1/3$: $\left( \frac{d^2y}{dx^2} + x^{1/4} \right)^3 = -\frac{dy}{dx}$. Expanding the left side,the highest derivative term is $\left( \frac{d^2y}{dx^2} \right)^3$. The order of the highest derivative is $2$. The degree is the power of the highest derivative after making the equation a polynomial in derivatives,which is $3$. Thus,the order is $2$ and the degree is $3$.

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

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