JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Medium

1 of 1

Analyzing +4 | -1

The differential equation whose solution is

{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {a^2}

is (

a

is a constant)

View answer and explanation

Correct option: B

{\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]^3} = {a^2}{\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^2}

Option B (

{\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]^3} = {a^2}{\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^2}

) is correct because it is the differential equation obtained by differentiating the given equation twice and eliminating the arbitrary constants h and k.

More Questions from DIFFERENTIAL EQUATIONS

Practice All Chapter MCQs →

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)