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JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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Let y = y(x) be the solution of the differential equation ex1−y2dx+(yx)dy =0,y(1)=−1.{e^x}\sqrt {1 - \mathrm{ }{y^2}} \mathrm{ }dx\mathrm{ } + \left( {\frac{y}{x}} \right)dy\, = \mathrm{ }0,\mathrm{ }y\left( 1 \right)\mathrm{ } = \mathrm{ } - 1. Then, the value of [y(3)]2{\left[ {y\left( 3 \right)} \right]^2} is equal to
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Correct option: B
1−4e6\mathrm{1} - \mathrm{ 4}{\mathrm{e}^6}
Option B (1−4e6\mathrm{1} - \mathrm{ 4}{\mathrm{e}^6}) is correct because solving the differential equation with the given initial condition and then evaluating at x=3 leads to this result.

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Let y=y(x)y=y(x) be the solution of the differential equation (3y2−5x2)ydx+2x(x2−y2)dy=0\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)=1y(1)=1. Then ∣(y(2))3−12y(2)∣\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 xx-axis, are If x2+y2=1{x^2} + {y^2} = 1, then(y′=dydx,y′′=d2ydx2)\left( {y' = \frac{{dy}}{{dx}},y'' = \frac{{{d^2}y}}{{d{x^2}}}} \right) If x=Acos⁡4t+Bsin⁡4t,x = A\cos 4t + B\sin 4t, then d2xdt2\frac{{{d^2}x}}{{d{t^2}}} is equal to Let y=y1(x)y=y_{1}(x) and y=y2(x)y=y_{2}(x) be the solution curves of the differential equation dydx=y+7\frac{d y}{d x}=y+7 with initial conditions y1(0)=0y_{1}(0)=0 and y2(0)=1y_{2}(0)=1 respectively. Then the curves y=y1(x)y=y_{1}(x) and y=y2(x)y=y_{2}(x) intersect at Let y=y(t)y=y(t) be a solution of the differential equation dydt+αy=γe−βt\frac{d y}{d t}+\alpha y=\gamma e^{-\beta t}
where, α>0,β>0\alpha>0, \beta>0 and γ>0\gamma>0. Then lim⁡x→∞y(t)\lim \limits_{{x \rightarrow \infty}} y(t)
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