The order and degree of the differential equa | vedclass

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(C) Given the differential equation: $\{1+(\frac{dy}{dx})^2\}^{\frac{3}{2}}=\frac{d^2y}{dx^2}$.To find the degree,we must eliminate the fractional exp

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) Given the differential equation: $\{1+(\frac{dy}{dx})^2\}^{\frac{3}{2}}=\frac{d^2y}{dx^2}$.To find the degree,we must eliminate the fractional exponent by squaring both sides:$\{1+(\frac{dy}{dx})^2\}^3 = (\frac{d^2y}{dx^2})^2$.The highest order derivative present is $\frac{d^2y}{dx^2}$,so the order $p = 2$.The power of the highest order derivative after making the equation a polynomial in derivatives is $2$,so the degree $q = 2$.Therefore,$p+q = 2+2 = 4$.

The order and degree of the differential equation $\{1+(\frac{dy}{dx})^2\}^{\frac{3}{2}}=\frac{d^2y}{dx^2}$ are $p$ and $q$ respectively. Then,$p+q=$ . . . . . . .

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