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1. ⇒ (MHT CET 2021 21th September Evening Shift )

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

A. $m = 3, n = 1$

B. $m = 3, n = 2$

C. $m = 3, n = 3$

D. $m = 3, n = 5$

Correct Option is (B)

$\begin{aligned} We have {(\frac{d^{2} y}{d x^{2}})}^{5} + \frac{4 (\frac{d^{2} y}{d x^{2}})}{(\frac{d^{3} y}{d x^{3}})} + (\frac{d^{3} y}{d x^{3}}) = x^{2} - 1 \\ ∴ (\frac{d^{3} y}{d x^{3}}) {(\frac{d^{2} y}{d x^{2}})}^{5} + 4 (\frac{d^{2} y}{d x^{2}}) + {(\frac{d^{3} y}{d x^{3}})}^{2} = (x^{2} - 1) (\frac{d^{3} y}{d x^{3}}) \\ ∴ order = 3, degree = 2 \end{aligned}$

2. ⇒ (MHT CET 2021 20th September Evening Shift )

If $m$ is order and $n$ is degree of the differential equation $y = \frac{d p}{d x} + \sqrt{a^{2} p^{2} - b^{2}}$ , where $p = \frac{d y}{d x}$ , then the value of $m + n$ is

A. 2

B. 3

C. 4

D. 5

Correct Option is (C)

$\begin{aligned} y = \frac{d p}{d x} + \sqrt{a^{2} p^{2} - b^{2}} = \frac{d}{d x} (\frac{d y}{d x}) + \sqrt{a^{2} {(\frac{d y}{d x})}^{2} - b^{2}} \\ ∴ y - \frac{d^{2} y}{d x^{2}} = \sqrt{a^{2} {(\frac{d y}{d x})}^{2} - b^{2}} \end{aligned}$

Squaring both sides, we write

$\begin{aligned} y^{2} + {(\frac{d^{2} y}{d x})}^{2} - 2 y \frac{d^{2} y}{d x^{2}} = a^{2} {(\frac{d y}{d x})}^{2} - b^{2} \\ ∴ order = 2 and degree = 2 \end{aligned}$

⇒Defferential Equation