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(C) Given differential equation: $\frac{y^2 e^{-1/y}}{\sqrt{x}} dx = 2 \sec \sqrt{x} dy$. Separate the variables: $\frac{dx}{\sqrt{x} \sec \sqrt{x}} =

Vedclass · Accepted Answer

$\sin \sqrt{x} + e^{1/y} = e$

Vedclass · Answer

(C) Given differential equation: $\frac{y^2 e^{-1/y}}{\sqrt{x}} dx = 2 \sec \sqrt{x} dy$. Separate the variables: $\frac{dx}{\sqrt{x} \sec \sqrt{x}} = \frac{2 dy}{y^2 e^{-1/y}}$. This simplifies to: $\cos \sqrt{x} \frac{dx}{\sqrt{x}} = 2 e^{1/y} \frac{dy}{y^2}$. Integrate both sides: $\int \cos \sqrt{x} \frac{dx}{\sqrt{x}} = \int 2 e^{1/y} \frac{dy}{y^2}$. Let $u = \sqrt{x}$, then $du = \frac{1}{2\sqrt{x}} dx \implies \frac{dx}{\sqrt{x}} = 2 du$. Let $v = 1/y$, then $dv = -\frac{1}{y^2} dy \implies \frac{dy}{y^2} = -dv$. Substituting these: $\int \cos(u) (2 du) = \int 2 e^v (-dv)$. $2 \sin(u) = -2 e^v + C$. $\sin \sqrt{x} = -e^{1/y} + C' \implies \sin \sqrt{x} + e^{1/y} = C'$. The curve passes through $(\pi^2, 1)$, so $\sin \sqrt{\pi^2} + e^{1/1} = C' \implies \sin \pi + e = C' \implies 0 + e = C'$. Thus, $C' = e$. The equation is $\sin \sqrt{x} + e^{1/y} = e$.

Let $S$ be the family of curves given by the general solution of the differential equation $\frac{y^2 e^{-1 / y}}{\sqrt{x}} dx - 2 \sec \sqrt{x} dy = 0$. Then the equation of the curve belonging to $S$ and passing through $(\pi^2, 1)$ is

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