JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Easy

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Application +4 | -1

y(x)

satisfies the differential equation

y^{\prime}-y \tan x=2 x \sec x

and

y(0)=0

, then

View answer and explanation

Correct option: A

y\left(\frac{\pi}{4}\right)=\frac{\pi^2}{8 \sqrt{2}}, y^{\prime}\left(\frac{\pi}{3}\right)=\frac{4 \pi}{3}+\frac{2 \pi^2}{3 \sqrt{3}}

Option A (

y(\frac{{\pi}}{{4}})=\frac{{\pi^2}}{{8 \sqrt{{2}}}}, y^{\prime}(\frac{{\pi}}{{3}})=\frac{{4 \pi}}{{3}}+\frac{{2 \pi^2}}{{3 \sqrt{{3}}}}

) is correct because it provides the correct values of y and y' at the specified points, obtained by solving the given differential equation with the initial condition y(0) = 0.

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)