1
MHT CET 2025 20th April Evening Shift
MCQ (Single Correct Answer)
+2
-0

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

A
$\quad \cos \left(\frac{7}{x}\right)=1$
B
$\quad \cos \left(\frac{y}{x-1}\right)=7$
C
$\quad \cos \left(\frac{y-1}{x}\right)=7$
D
None
2
MHT CET 2025 20th April Evening Shift
MCQ (Single Correct Answer)
+2
-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

A
$2 x \mathrm{e}^{\tan ^{-1} y}=\mathrm{e}^{2 \tan ^{-1} y}+\mathrm{k}$, where k is the constant of integration
B
$x \cdot \mathrm{e}^{\tan ^{-1} y}=\mathrm{e}^{\tan ^{-1} y}+\mathrm{k}$, where k is the constant of integration
C
$x \cdot \mathrm{e}^{2 \tan ^{-1} y}=\mathrm{e}^{\tan ^{-1} y}+\mathrm{k}$, where k is the constant of integration
D
$\quad x=2+\mathrm{k} \cdot \mathrm{e}^{-\tan ^{-1} y}$, where k is the constant of integration
3
MHT CET 2025 20th April Evening Shift
MCQ (Single Correct Answer)
+2
-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

A
3 years
B
5 years
C
4 years
D
6 years
4
MHT CET 2025 20th April Morning Shift
MCQ (Single Correct Answer)
+2
-0

The general solution of

$x(x-1) \frac{\mathrm{d} y}{\mathrm{~d} x}=x^3(2 x-1)+(x-2) y$ is

A
$y(x-1)=x^3+\mathrm{c}(x-1)$, where c is the constant of integration.
B
$y=x^3(x-1)+\mathrm{c}$, where c is the constant of integration.
C
$y(x-1)=x^3(x-1)+\mathrm{cx}^2$, where c is the constant of integration.
D
$y(x-1)=x^3(x-1)+\mathrm{c}$, where c is the constant of integration.
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