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JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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Application +4 | -1

A wet porous substance in the open air loses its moisture at a rate proportional to the moisture content. If a sheet hung in the wind loses half its moisture during the first hour, then the time when it would have lost

99.9 \%

of its moisture is (weather conditions remaining same)

View answer and explanation

Correct option: C

approximately

10 \mathrm{~h}

Option C (approximately 10 h) is correct because it is derived from solving the differential equation and applying the given conditions to find the time when 99.9% of the moisture is lost.

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y=y(x)

be the solution of the differential equation

\left(3 y^{2}-5 x^{2}\right) y d x+2 x\left(x^{2}-y^{2}\right) d y=0

y(1)=1

\left|(y(2))^{3}-12 y(2)\right|

is equal to : The degree and order of the differential equation whose solution is a parabola whose axis is

x

{x^2} + {y^2} = 1

\left( {y' = \frac{{dy}}{{dx}},y'' = \frac{{{d^2}y}}{{d{x^2}}}} \right)

x = A\cos 4t + B\sin 4t,

\frac{{{d^2}x}}{{d{t^2}}}

is equal to Let

y=y_{1}(x)

y=y_{2}(x)

be the solution curves of the differential equation

\frac{d y}{d x}=y+7

with initial conditions

y_{1}(0)=0

y_{2}(0)=1

respectively. Then the curves

y=y_{1}(x)

y=y_{2}(x)

intersect at Let

y=y(t)

be a solution of the differential equation

\frac{d y}{d t}+\alpha y=\gamma e^{-\beta t}

\alpha>0, \beta>0

\gamma>0

\lim \limits_{{x \rightarrow \infty}} y(t)