If m is the order and n is the degree of the | vedclass

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

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$m=3, n=2$

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(B) Given the differential equation: $\left(\frac{d^2 y}{d x^2}\right)^5+4 \frac{\left(\frac{d^2 y}{d x^2}\right)}{\left(\frac{d^3 y}{d x^3}\right)}+\left(\frac{d^3 y}{d x^3}\right)=x^2-1$.Multiply the entire equation by $\left(\frac{d^3 y}{d x^3}\right)$ to eliminate the fraction:$\left(\frac{d^3 y}{d x^3}\right) \left(\frac{d^2 y}{d x^2}\right)^5 + 4 \left(\frac{d^2 y}{d x^2}\right) + \left(\frac{d^3 y}{d x^3}\right)^2 = (x^2-1) \left(\frac{d^3 y}{d x^3}\right)$.The highest order derivative present is $\frac{d^3 y}{d x^3}$,so the order $m = 3$.The highest power of the highest order derivative after clearing fractions is $2$,so the degree $n = 2$.Therefore,$m=3$ and $n=2$.

If $m$ is the order and $n$ is the degree of the differential equation $\left(\frac{d^2 y}{d x^2}\right)^5+4 \frac{\left(\frac{d^2 y}{d x^2}\right)}{\left(\frac{d^3 y}{d x^3}\right)}+\left(\frac{d^3 y}{d x^3}\right)=x^2-1$,then:

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