If 'm' and 'n' are the order and degree of the differential equation $(y^{''})^{5} + 4 \cdot \frac{(y^{''})^{3}}{y^{'''}} + y^{'''}$ .

m = 3, n = 5

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m = 3, n = 1

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m = 3, n = 3

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m = 3, n = 2

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Solution

The correct option is D $m = 3, n = 2$
$(y^{''})^{5} + 4. \frac{(y^{''})^{3}}{y^{'''}} + y^{'''}$
In the standard form, it can be written as $\frac{(y^{''})^{5} y^{'''} + 4. (y^{''})^{3} + (y^{'''})^{2}}{y^{'''}}$
The order of a differential equation is defined as the highest order derivative of the dependent variable with respect to the independent variable.
Here, order thus is 3 since we have third differential as the highest.
Degree is defined as the highest power of the highest order derivative, which is the power of $y^{'''}$ , which is $2$
Hence $m = 3, n = 2$

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{(y_{2})}^{5} + \frac{4 {(y_{2})}^{3}}{y_{3}} + y_{3} = x^{2} - 1

If ‘m’ and ‘n’ are the order and degree of the differential equation $(y^{''})^{5} + \frac{4 (y^{''})^{3}}{y^{'''}} + y^{'''} = s i n x$ , then

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{(\frac{d^{2} y}{d x^{2}})}^{5} + 4 \frac{(\frac{d^{2} y}{d x^{2}})}{(\frac{d^{3} y}{d x^{3}})} + (\frac{d^{3} y}{d x^{3}}) = x^{2} - 1

(m, n) =

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