If m and n are the order and degree of the di | vedclass

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(B) The given differential equation is $\left(\frac{d^{2} y}{d x^{2}}\right)^{5}+4 \cdot \frac{\left(\frac{d^{2} y}{d x^{2}}\right)^{3}}{\left(\frac{d

Vedclass · Accepted Answer

$m=3, n=2$

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(B) The given differential equation is $\left(\frac{d^{2} y}{d x^{2}}\right)^{5}+4 \cdot \frac{\left(\frac{d^{2} y}{d x^{2}}\right)^{3}}{\left(\frac{d^{3} y}{d x^{3}}\right)}+\left(\frac{d^{3} y}{d x^{3}}\right)=x^{2}-1$.To find the order and degree,we first eliminate the fraction by multiplying the entire equation by $\frac{d^{3} y}{d x^{3}}$:$\left(\frac{d^{2} y}{d x^{2}}\right)^{5} \cdot \left(\frac{d^{3} y}{d x^{3}}\right) + 4 \left(\frac{d^{2} y}{d x^{2}}\right)^{3} + \left(\frac{d^{3} y}{d x^{3}}\right)^{2} = (x^{2}-1) \left(\frac{d^{3} y}{d x^{3}}\right)$.The highest order derivative present in the equation is $\frac{d^{3} y}{d x^{3}}$,so the order $m = 3$.The degree is the highest power of the highest order derivative after the equation is made a polynomial in derivatives. Here,the highest power of $\frac{d^{3} y}{d x^{3}}$ is $2$,so the degree $n = 2$.

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

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