The order and the degree of the differential | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) To find the order and degree, we must first eliminate the radicals by raising both sides to the power of $6$ (the least common multiple of $2$ and

Vedclass · Accepted Answer

$2, 6$

Vedclass · Answer

(D) To find the order and degree, we must first eliminate the radicals by raising both sides to the power of $6$ (the least common multiple of $2$ and $3$).Given equation: $(1 + (y'')^2)^{1/2} = (x + (y')^3)^{1/3}$.Raising both sides to the power of $6$:$((1 + (y'')^2)^{1/2})^6 = ((x + (y')^3)^{1/3})^6$$(1 + (y'')^2)^3 = (x + (y')^3)^2$.Expanding the left side: $1 + 3(y'')^2 + 3(y'')^4 + (y'')^6 = (x + (y')^3)^2$.The highest order derivative present is $y'' = \frac{d^2y}{dx^2}$, so the order is $2$.The exponent of the highest order derivative after rationalizing the equation is $6$. Therefore, the degree is $6$.

The order and the degree of the differential equation $\sqrt{1 + (\frac{d^2y}{dx^2})^2} = \sqrt[3]{x + (\frac{dy}{dx})^3}$ are respectively . . . . . . and . . . . . . .

Similar Questions

The number of arbitrary constants in the particular solution of a differential equation of third order is:

The order and degree of the differential equation $(\frac{d^3y}{dx^3})^{\frac{2}{3}} - 3\frac{d^2y}{dx^2} + 5\frac{dy}{dx} + 4 = 0$ are respectively:

The order and degree of the differential equation $3 - (\frac{d^3y}{dx^3})^{7/3} = (\frac{dy}{dx})^5$ are respectively

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

Determine the order and degree (if defined) of the differential equation $y'' + (y')^2 + 2y = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module