1
MHT CET 2024 3rd May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The general solution of the differential equation $x \cos y \mathrm{~d} y=\left(x \mathrm{e}^{\mathrm{x}} \log x+\mathrm{e}^x\right) \mathrm{d} x$ is given by

A
$\sin y=\mathrm{e}^x+\operatorname{clog} x$, where c is a constant of integration.
B
$\sin y=\mathrm{e}^{\mathrm{x}} \log x+\mathrm{c}$, where c is a constant of integration.
C
$\mathrm{e}^x \sin y=\log x+\mathrm{c}$, where c is a constant of integration.
D
$\sin y=\mathrm{ce}^x+\log x$, where c is a constant of integration.
2
MHT CET 2024 3rd May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If order and degree of the differential equation $\left(\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}\right)^5+4 \frac{\left(\frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}\right)^5}{\left(\frac{\mathrm{~d}^3 y}{\mathrm{~d} x^3}\right)}+\frac{\mathrm{d}^3 y}{\mathrm{~d} x^3}=\sin x$, are $m$ and $n$ respectively, then the value of $\left(\mathrm{m}^2+\mathrm{n}^2\right)$ is equal to

A
29
B
13
C
5
D
8
3
MHT CET 2024 3rd May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If a body cools from $80^{\circ} \mathrm{C}$ to $60^{\circ} \mathrm{C}$ in the room temperature of $30^{\circ} \mathrm{C}$ in 30 min , then the temperature of a body after one hour is

A
$42^{\circ} \mathrm{C}$
B
$24^{\circ} \mathrm{C}$
C
$48^{\circ} \mathrm{C}$
D
$56^{\circ} \mathrm{C}$
4
MHT CET 2024 3rd May Morning Shift
MCQ (Single Correct Answer)
+2
-0

The differential equation $\left[\frac{1+\left(\frac{d y}{d x}\right)^2}{\left(\frac{d^2 y}{d x^2}\right)}\right]^{\frac{3}{2}}=\mathrm{kx}$ is of

A
order $=2$, degree $=3$
B
order $=3$, degree $=2$
C
order $=2$, degree $=2$
D
order $=3$, degree $=3$
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