The order and degree of the differential equa | vedclass

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(C) Given the differential equation: $(\frac{d^3y}{dx^3}+y\frac{dy}{dx})^{\frac{7}{5}} =x^3\frac{d^2y}{dx^2}$.To find the degree,we must eliminate the

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(C) Given the differential equation: $(\frac{d^3y}{dx^3}+y\frac{dy}{dx})^{\frac{7}{5}} =x^3\frac{d^2y}{dx^2}$.To find the degree,we must eliminate the fractional exponent by raising both sides to the power of $5$:$(\frac{d^3y}{dx^3}+y\frac{dy}{dx})^7 = (x^3\frac{d^2y}{dx^2})^5$.The order $m$ is the highest derivative present,which is $\frac{d^3y}{dx^3}$,so $m = 3$.The degree $n$ is the power of the highest derivative after the equation is expressed as a polynomial in derivatives,which is $7$,so $n = 7$.Therefore,$m + n = 3 + 7 = 10$.

The order and degree of the differential equation $(\frac{d^3y}{dx^3}+y\frac{dy}{dx})^{\frac{7}{5}} =x^3\frac{d^2y}{dx^2}$ are $m$ and $n$ respectively,then $(m + n)$ is equal to-

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