JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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Analyzing +4 | -1

y:[1, \infty) \rightarrow[1, \infty)

is a function of

\mathrm{x}

such that

2 x=y^{1 / 5}+y^{-1 / 5}

and

\left(x^2-1\right) \frac{d^2 y}{d x^2}+x \frac{d y}{d x}=k y(k \in R)

Then

\mathrm{k}=

View answer and explanation

Correct option: C

The correct answer is 25 because after differentiating the given equation and simplifying, the value of k in the differential equation is found to be 25.

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)