MHT CET 2025 21st April Morning Shift | Differential Equations Question 43 | Mathematics

The assets of a person reduced in his business such that the rate of reduction is proportional to the square root of the existing assets. If the assets were initially ₹ 10 lakhs and due to loss they reduce to ₹ 10000 after 3 years, then the number of years required for the person to be bankrupt will be

$\frac{20}{3}$ years

$\frac{10}{3}$ years

$\frac{10}{9}$ years

$\frac{20}{9}$ years

MHT CET 2025 20th April Evening Shift

MCQ (Single Correct Answer)

-0

If the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}+\frac{x}{y}=\frac{\mathrm{a}}{y}$ where a is constant, represents a family of circles then the radius of the circle is $\qquad$

$\mathrm{a}+2 \mathrm{c}$, where c is the constant of integration

$\sqrt{\mathrm{a}^2+2 \mathrm{c}}$, where c is the constant of integration

$\mathrm{a}^2+2 \mathrm{c}$, where c is the constant of integration

$\sqrt{a+c}$, where $c$ is the constant of integration

MHT CET 2025 20th April Evening Shift

MCQ (Single Correct Answer)

-0

The particular solution of the differential equation $\cos \left(\frac{d y}{d x}\right)=7, y=1$ at $x=0$ is

$\quad \cos \left(\frac{7}{x}\right)=1$

$\quad \cos \left(\frac{y}{x-1}\right)=7$

$\quad \cos \left(\frac{y-1}{x}\right)=7$

None

MHT CET 2025 20th April Evening Shift

MCQ (Single Correct Answer)

-0

The solution of $\left(1+y^2\right)+\left(x-\mathrm{e}^{\tan ^{-1} y}\right) \frac{\mathrm{d} y}{\mathrm{~d} x}=0$ is

$2 x \mathrm{e}^{\tan ^{-1} y}=\mathrm{e}^{2 \tan ^{-1} y}+\mathrm{k}$, where k is the constant of integration

$x \cdot \mathrm{e}^{\tan ^{-1} y}=\mathrm{e}^{\tan ^{-1} y}+\mathrm{k}$, where k is the constant of integration

$x \cdot \mathrm{e}^{2 \tan ^{-1} y}=\mathrm{e}^{\tan ^{-1} y}+\mathrm{k}$, where k is the constant of integration

$\quad x=2+\mathrm{k} \cdot \mathrm{e}^{-\tan ^{-1} y}$, where k is the constant of integration

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