Let y=y(x) be a solution curve of the differe | vedclass

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(A) The given differential equation is $(1-x^2 y^2) dx = y dx + x dy$.We know that $d(xy) = y dx + x dy$. Substituting this into the equation,we get $

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

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(A) The given differential equation is $(1-x^2 y^2) dx = y dx + x dy$.We know that $d(xy) = y dx + x dy$. Substituting this into the equation,we get $(1-(xy)^2) dx = d(xy)$.Rearranging the terms,we have $dx = \frac{d(xy)}{1-(xy)^2}$.Integrating both sides,$\int dx = \int \frac{d(xy)}{1-(xy)^2}$.Using the formula $\int \frac{du}{1-u^2} = \frac{1}{2} \ln \left| \frac{1+u}{1-u} \right| + C$,we get $x = \frac{1}{2} \ln \left| \frac{1+xy}{1-xy} \right| + C$.Given $y(1) = 2$,substituting $x=1$ and $y=2$ gives $1 = \frac{1}{2} \ln \left| \frac{1+2}{1-2} \right| + C$,so $1 = \frac{1}{2} \ln 3 + C$,which implies $C = 1 - \frac{1}{2} \ln 3$.Now,for $x=2$ and $y=\alpha$,we have $2 = \frac{1}{2} \ln \left| \frac{1+2\alpha}{1-2\alpha} \right| + 1 - \frac{1}{2} \ln 3$.$1 + \frac{1}{2} \ln 3 = \frac{1}{2} \ln \left| \frac{1+2\alpha}{1-2\alpha} \right|$,which simplifies to $2 + \ln 3 = \ln \left| \frac{1+2\alpha}{1-2\alpha} \right|$.Taking exponential on both sides,$3e^2 = \left| \frac{1+2\alpha}{1-2\alpha} \right|$.Case $1$: $\frac{1+2\alpha}{1-2\alpha} = 3e^2 \implies 1+2\alpha = 3e^2 - 6e^2\alpha \implies \alpha(2+6e^2) = 3e^2-1 \implies \alpha = \frac{3e^2-1}{2(3e^2+1)}$.Case $2$: $\frac{1+2\alpha}{1-2\alpha} = -3e^2 \implies 1+2\alpha = -3e^2 + 6e^2\alpha \implies \alpha(2-6e^2) = -3e^2-1 \implies \alpha = \frac{3e^2+1}{2(3e^2-1)}$.

Let $y=y(x)$ be a solution curve of the differential equation,$(1-x^2 y^2) dx = y dx + x dy$. 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

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