MHT CET 2024 9th May Morning Shift | Differential Equations Question 29 | Mathematics

If $y=y(x)$ is the solution of the differential equation $x \frac{\mathrm{dy}}{\mathrm{d} x}+2 y=x^2$ satisfying $y(1)=1$, then the value of $y\left(\frac{1}{2}\right)$ is

$\frac{7}{64}$

$\frac{1}{4}$

$\frac{13}{6}$

$\frac{49}{16}$

MHT CET 2024 9th May Morning Shift

MCQ (Single Correct Answer)

-0

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)

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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)

-0

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$

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