The order and degree of the differential equa | vedclass

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(C) Given the differential equation: $1+(\frac{dy}{dx})^2=\sqrt{\frac{d^2y}{dx^2}}$To find the order and degree,we first eliminate the radical by squa

Vedclass · Accepted Answer

$2, 1$

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(C) Given the differential equation: $1+(\frac{dy}{dx})^2=\sqrt{\frac{d^2y}{dx^2}}$To find the order and degree,we first eliminate the radical by squaring both sides:$(1+(\frac{dy}{dx})^2)^2 = \frac{d^2y}{dx^2}$The highest order derivative present in the equation is $\frac{d^2y}{dx^2}$,so the order is $2$.The power of the highest order derivative after making the equation a polynomial in derivatives is $1$,so the degree is $1$.Therefore,the order and degree are $2$ and $1$ respectively.

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

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