MHT CET 2024 15th May Morning Shift | Differential Equations Question 11 | Mathematics

A radio active substance has half-life of $h$ days, then its initial decay rate is given by Note that at $\mathrm{t}=0, \mathrm{M}=\mathrm{m}_{\mathrm{o}}$

$\frac{\mathrm{m}_{\mathrm{o}}}{\mathrm{h}}(\log 2)$

$\left(\mathrm{m}_{\mathrm{o}} \mathrm{h}\right)(\log 2)$

$-\frac{\mathrm{m}_{\mathrm{o}}}{\mathrm{h}}(\log 2)$

$\left(-\mathrm{m}_{\mathrm{o}} \mathrm{h}\right)(\log 2)$

MHT CET 2024 15th May Morning Shift

MCQ (Single Correct Answer)

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The differential equation of $y=\mathrm{e}^x\left(\mathrm{a}+\mathrm{bx}+x^2\right)$ is

$\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}+2 \frac{\mathrm{~d} y}{\mathrm{~d} x}-2 y=0$

$\frac{\mathrm{d}^2 y}{\mathrm{dx}^2}-2 \frac{\mathrm{~d} y}{\mathrm{~d} x}+y=0$

$\frac{\mathrm{d}^2 y}{\mathrm{dx}^2}-2 \frac{\mathrm{~d} y}{\mathrm{~d} x}-2 \mathrm{e}^x+y=0$

$\frac{\mathrm{d}^2 y}{\mathrm{dx}}+2 \frac{\mathrm{~d} y}{\mathrm{~d} x}-\mathrm{e}^x+2 y=0$

MHT CET 2024 11th May Evening Shift

MCQ (Single Correct Answer)

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An ice ball melts at the rate which is proportional to the amount of ice at that instant. Half of the quantity of ice melts in 15 minutes. $x_0$ is the initial quantity of ice. If after 30 minutes the amount of ice left is $\mathrm{kx}_0$, then the value of $k$ is

$\frac{1}{2}$

$\frac{1}{3}$

$\frac{1}{4}$

$\frac{1}{8}$

MHT CET 2024 11th May Evening Shift

MCQ (Single Correct Answer)

-0

Let $y=y(x)$ be the solution of the differential equation $x \frac{\mathrm{~d} y}{\mathrm{~d} x}+y=x \log x,(x>1)$ If $2(y(2))=\log 4-1$ then the value of $y(\mathrm{e})$ is

$\frac{\mathrm{e}^2}{4}$

$\frac{-\mathrm{e}^2}{2}$

$\frac{-\mathrm{e}}{2}$

$\frac{\mathrm{e}}{4}$

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