JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Easy

1 of 1

Application +4 | -1

The area of the region enclosed between the two circles

x^2+y^2=1

and

(x-1)^2+y^2=1

, is

View answer and explanation

Correct option: A

\left(\frac{2 \pi}{3}-\sqrt{\frac{3}{4}}\right)

Option A is correct because it represents the area of the intersection of two circles, calculated using geometry and trigonometry.

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y=y(x)

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

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x

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

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

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

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

y_{1}(0)=0

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y=y_{2}(x)

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)