Let $y=y(x)$ be a solution curve of the differential equation. $\left(1-x^2 y^2\right) d x=y d x+x d y$ If the line $x=1$ intersects the curve $y=y(x)$ at $y=2$ and the line $x=2$ intersects the curve $y=y(x)$ at $y=\alpha$, then a value of $\alpha$ is

$\frac{3 e^2}{2\left(3 e^2-1\right)}$

$\frac{3 e^2}{2\left(3 e^2+1\right)}$

JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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Application +4 | -1

Let

y=y(x)

be a solution curve of the differential equation.

\left(1-x^2 y^2\right) d x=y d x+x d y

If the line

x=1

intersects the curve

y=y(x)

y=2

and the line

x=2

intersects the curve

y=y(x)

y=\alpha

, then a value of

\alpha

View answer and explanation

Correct option: B

\frac{1+3 e^2}{2\left(3 e^2-1\right)}

Option B (

\frac{{1+3 e^2}}{{2(3 e^2-1)}}}

) is correct because it represents the value of

\alpha

obtained by solving the given differential equation and applying the initial conditions.

Practice All Chapter MCQs →

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)