JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Easy

1 of 1

Analyzing +4 | -1

A particle starts at the origin and moves along the

x

-axis in such a way that its velocity at the point

\left( {x,0} \right)

is given by the formula

\frac{{dx}}{{dt}} = {\cos ^2}\pi x

. Then, the particle never reaches the point on

View answer and explanation

Correct option: C

x = \frac{1}{2}

Option C (

x = \frac{1}{2}

) is correct because the velocity becomes zero at this point, implying the particle stops and never reaches it.

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y=y(x)

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

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

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

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

y_{1}(0)=0

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y=y(t)

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

\alpha>0, \beta>0

\gamma>0

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