JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Easy

1 of 1

Analyzing +4 | -1

The slope of the tangent to a curve

C: y=y(x)

at any point

(x, y)

on it is

\frac{2 e^{2 x}-6 e^{-x}+9}{2+9 e^{-2 x}}

. If

C

passes through the points

\left(0, \frac{1}{2}+\frac{\pi}{2 \sqrt{2}}\right)

and

\left(\alpha, \frac{1}{2} e^{2 \alpha}\right)

, then

e^\alpha

is equal to

View answer and explanation

Correct option: B

\frac{3}{\sqrt{2}}\left(\frac{3+\sqrt{2}}{3-\sqrt{2}}\right)

Option B is correct. The value of

e^\alpha

is found by solving the differential equation and using the given points to find the constant of integration and then solving for

e^\alpha

More Questions from DIFFERENTIAL EQUATIONS

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y=y(x)

\left(3 y^{2}-5 x^{2}\right) y d x+2 x\left(x^{2}-y^{2}\right) d y=0

y(1)=1

\left|(y(2))^{3}-12 y(2)\right|

x

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

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

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

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

y=y_{1}(x)

y=y_{2}(x)

\frac{d y}{d x}=y+7

y_{1}(0)=0

y_{2}(0)=1

y=y_{1}(x)

y=y_{2}(x)

y=y(t)

\frac{d y}{d t}+\alpha y=\gamma e^{-\beta t}

\alpha>0, \beta>0

\gamma>0

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