MHT CET 2021 20th September Morning Shift | Differential Equations Question 53 | Mathematics

A differential equation for the temperature 'T' of a hot body as a function of time, when it is placed in a bath which is held at a constant temperature of 32$$^\circ$$ F, is given by (where k is a constant of proportionality)

$$\mathrm{\frac{dT}{dt}=kT-32}$$

$$\mathrm{\frac{dT}{dt}=kT+32}$$

$$\mathrm{\frac{dT}{dt}=-k(T-32)}$$

$$\mathrm{\frac{dT}{dt}=32kT}$$

MHT CET 2021 20th September Morning Shift

MCQ (Single Correct Answer)

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The general solution of the differential equation $$\frac{d y}{d x}=\frac{x+y+1}{x+y-1}$$ is given by

$$y=x \log (x+y)+c$$

$$x-y=\log (x+y)+c$$

$$x+y=\log (x+y)+c$$

$$y=x+\log (x+y)+c$$

MHT CET 2021 20th September Morning Shift

MCQ (Single Correct Answer)

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The general solution of the differential equation $$x+y \frac{d y}{d x}=\sec \left(x^2+y^2\right)$$ is

$$\sin \left(x^2+y^2\right)=2 x+c$$

$$\sin \left(x^2+y^2\right)+2 x=c$$

$$\sin \left(x^2+y^2\right)+x=c$$

$$\cos \left(x^2+y^2\right)=2 x+c$$

MHT CET 2021 20th September Morning Shift

MCQ (Single Correct Answer)

-0

The differential equation of all circles which pass through the origin and whose centre lie on Y-axis is

$$\left(x^2-y^2\right) \frac{d y}{d x}-2 x y=0$$

$$\left(x^2+y^2\right) \frac{d y}{d x}-2 x y=0$$

$$\left(x^2+y^2\right) \frac{d y}{d x}+2 x y=0$$

$$\left(x^2-y^2\right) \frac{d y}{d x}+2 x y=0$$

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