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

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Question

(D) The given differential equation is $\left(y^{\prime \prime}\right)^{5}+4 \cdot \frac{\left(y^{\prime \prime}\right)^{3}}{y^{\prime \prime \prime}}

Vedclass · Accepted Answer

$m=3, n=2$

Vedclass · Answer

(D) The given differential equation is $\left(y^{\prime \prime}\right)^{5}+4 \cdot \frac{\left(y^{\prime \prime}\right)^{3}}{y^{\prime \prime \prime}}+y^{\prime \prime \prime}=\sin x$. To find the order and degree,we must eliminate the fraction by multiplying the entire equation by $y^{\prime \prime \prime}$. Multiplying by $y^{\prime \prime \prime}$,we get: $\left(y^{\prime \prime}\right)^{5} \cdot y^{\prime \prime \prime} + 4 \cdot \left(y^{\prime \prime}\right)^{3} + \left(y^{\prime \prime \prime}\right)^{2} = \sin x \cdot y^{\prime \prime \prime}$. The highest order derivative present is $y^{\prime \prime \prime}$,so the order $m = 3$. The degree $n$ is the power of the highest order derivative after the equation is expressed as a polynomial in derivatives. The term with the highest order derivative is $\left(y^{\prime \prime \prime}\right)^{2}$,so the degree $n = 2$.

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

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