MHT CET 2024 9th May Morning Shift | Differential Equations Question 30 | Mathematics

The curve satisfying the differential equation $y \mathrm{~d} x-\left(x+3 y^2\right) \mathrm{dy}=0$ and passing through the point $(1,1)$ also passes through the point

$\left(\frac{1}{4}, \frac{1}{2}\right)$

$\left(\frac{1}{4},-\frac{1}{2}\right)$

$\left(\frac{1}{3},-\frac{1}{3}\right)$

$\left(-\frac{1}{3}, \frac{1}{3}\right)$

MHT CET 2024 4th May Evening Shift

MCQ (Single Correct Answer)

-0

The general solution of the differential equation $\frac{1}{x} \frac{\mathrm{~d} y}{\mathrm{~d} x}=\tan ^{-1}$ is

$y+\frac{x^2 \tan ^{-1} x}{2}+\mathrm{c}=0$, where c is a constant of integration.

$y+x \tan ^{-1} x+\mathrm{c}=0$, where c is a constant integration.

$y-x-\tan ^{-1} x+\mathrm{c}=0$, where is a constant of integration.

$y=\frac{x^2 \tan ^{-1} x}{2}-\frac{1}{2}\left(x-\tan ^{-1} x\right)+\mathrm{c}$, where c is constant of integration.

MHT CET 2024 4th May Evening Shift

MCQ (Single Correct Answer)

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The differential equation obtained by eliminating arbitrary constant from the equation $y^2=(x+c)^3$ is

$\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^3=27 y$

$\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^3=-27 y$

$8\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^3=27 y$

$ 8\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^3+27 y=0$

MHT CET 2024 4th May Evening Shift

MCQ (Single Correct Answer)

-0

The decay rate of radium is proportional to the amount present at any time $t$. If initially 60 gms was present and half life period of radium is 1600 years, then the amount of radium present after 3200 years is

20 grams

15 grams

12 grams

10 grams

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