MHT CET 2023 14th May Evening Shift | Differential Equations Question 2 | Mathematics

If a body cools from $$80^{\circ} \mathrm{C}$$ to $$50^{\circ} \mathrm{C}$$ in the room temperature of $$25^{\circ} \mathrm{C}$$ in 30 minutes, then the temperature of the body after 1 hour is

$$31.36^{\circ} \mathrm{C}$$

$$32.25^{\circ} \mathrm{C}$$

$$36.36^{\circ} \mathrm{C}$$

$$33.25^{\circ} \mathrm{C}$$

MHT CET 2023 14th May Evening Shift

MCQ (Single Correct Answer)

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The differential equation representing the family of curves $$y^2=2 \mathrm{c}(x+\sqrt{\mathrm{c}})$$, where $$\mathrm{c}$$ is a positive parameter, is of

order 1, degree 4

order 2, degree 3

order 2, degree 4

order 1, degree 3

MHT CET 2023 14th May Morning Shift

MCQ (Single Correct Answer)

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General solution of the differential equation $$\cos x(1+\cos y) \mathrm{d} x-\sin y(1+\sin x) \mathrm{d} y=0$$ is

$$(1+\cos x)(1+\sin y)=\mathrm{c}$$, where $$\mathrm{c}$$ is a constant of integration.

$$1+\sin x+\cos y=\mathrm{c}$$, where $$\mathrm{c}$$ is a constant of integration.

$$(1+\sin x)(1+\cos y)=\mathrm{c}$$, where $$\mathrm{c}_{\text {is }}$$ constant of integration.

$$1+\sin x \cos y=\mathrm{c}$$, where $$\mathrm{c}$$ is a constant of integration.

MHT CET 2023 14th May Morning Shift

MCQ (Single Correct Answer)

-0

The differential equation of $$y=\mathrm{e}^x(\mathrm{a} \cos x+\mathrm{b} \sin x)$$ is

$$\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}-2 \frac{\mathrm{d} y}{\mathrm{~d} x}-y=0$$

$$\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}+2 \frac{\mathrm{d} y}{\mathrm{~d} x}+2 y=0$$

$$\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}-2 \frac{\mathrm{d} y}{\mathrm{~d} x}+y=0$$

$$\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}-2 \frac{\mathrm{d} y}{\mathrm{~d} x}+2 y=0$$

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