If the order and degree of the differential e | vedclass

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Question

(B) Given the differential equation: $\left(\frac{d^2 y}{dx^2}\right)^5 + 4 \frac{\left(\frac{d^2 y}{dx^2}\right)^5}{\frac{d^3 y}{dx^3}} + \frac{d^3 y

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(B) Given the differential equation: $\left(\frac{d^2 y}{dx^2}\right)^5 + 4 \frac{\left(\frac{d^2 y}{dx^2}\right)^5}{\frac{d^3 y}{dx^3}} + \frac{d^3 y}{dx^3} = \sin x$. Multiply the entire equation by $\frac{d^3 y}{dx^3}$ to eliminate the fraction: $\left(\frac{d^2 y}{dx^2}\right)^5 \cdot \frac{d^3 y}{dx^3} + 4\left(\frac{d^2 y}{dx^2}\right)^5 + \left(\frac{d^3 y}{dx^3}\right)^2 = \sin x \cdot \frac{d^3 y}{dx^3}$. The highest order derivative present is $\frac{d^3 y}{dx^3}$,so the order $m = 3$. The highest power of the highest order derivative is $2$,so the degree $n = 2$. Therefore,$m^2 + n^2 = 3^2 + 2^2 = 9 + 4 = 13$.

If the order and degree of the differential equation $\left(\frac{d^2 y}{dx^2}\right)^5 + 4 \frac{\left(\frac{d^2 y}{dx^2}\right)^5}{\left(\frac{d^3 y}{dx^3}\right)} + \frac{d^3 y}{dx^3} = \sin x$ are $m$ and $n$ respectively,then the value of $(m^2 + n^2)$ is equal to

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