JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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Application +4 | -1

If the area of the figure bounded by a curve, the

x

-axis, and two ordinates, one of which is constant, the other variable is equal to the ratio of the cube of the variable ordinate to the variable abscissa, then the curve is

View answer and explanation

Correct option: A

\left(2 y^2-x^2\right)^3=C x^2

Option A is correct because the given condition translates to a differential equation that, when solved, yields the equation (2y^2-x²)³=Cx².

More Questions from DIFFERENTIAL EQUATIONS

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Let

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

such that

y(1)=1

. Then

\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

-axis, are If

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

, then

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

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

then

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

is equal to Let

y=y_{1}(x)

and

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

and

y_{2}(0)=1

respectively. Then the curves

y=y_{1}(x)

and

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}

where,

\alpha>0, \beta>0

and

\gamma>0

. Then

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