JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Easy

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Application +4 | -1

Let the solution curve

y=y(x)

of the differential equation

\frac{d y}{d x}-\frac{3 x^5 \tan ^{-1}\left(x^3\right)}{\left(1+x^6\right)^{3 / 2}} y=2 x \exp \left\{\frac{x^3-\tan ^{-1} x^3}{\sqrt{\left(1+x^6\right)}}\right\}

pass through the origin. Then

y(1)

is equal to

View answer and explanation

Correct option: B

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

Option B (exp((4-π)/(4√2))) is correct because it is the value of y(1) obtained after solving the given differential equation and applying 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)