MHT CET 2025 20th April Evening Shift | Differential Equations Question 33 | Mathematics

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

MHT CET 2025 20th April Evening Shift

MCQ (Single Correct Answer)

-0

The rate of reduction of a persons assets is proportional to the square root of the existing assets. The assets reduced from 25 lakhs to 6.25 lakhs in 2 years. This rate of reduction of his assets will make him bankrupt in

3 years

5 years

4 years

6 years

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