Let x = x(y) be the solution of the different | vedclass

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(B) Given the differential equation: $y = (x - y \frac{dx}{dy}) \sin(\frac{x}{y})$.Rearranging the terms: $\frac{y}{\sin(x/y)} = x - y \frac{dx}{dy}$.

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$2(\ln 2)^2 - 1$

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(B) Given the differential equation: $y = (x - y \frac{dx}{dy}) \sin(\frac{x}{y})$.Rearranging the terms: $\frac{y}{\sin(x/y)} = x - y \frac{dx}{dy}$.This can be rewritten as: $y \csc(x/y) = x - y \frac{dx}{dy}$.Alternatively,consider the differential form: $y dy = (x dy - y dx) \sin(x/y)$.Dividing by $y^2$: $\frac{dy}{y} = \frac{x dy - y dx}{y^2} \sin(x/y)$.Note that $d(x/y) = \frac{y dx - x dy}{y^2}$,so $\frac{x dy - y dx}{y^2} = -d(x/y)$.Thus,$\frac{dy}{y} = -\sin(x/y) d(x/y)$.Integrating both sides: $\int \frac{1}{y} dy = -\int \sin(x/y) d(x/y)$.$\ln y = \cos(x/y) + C$.Using the condition $x(1) = \pi/2$: $\ln(1) = \cos(\frac{\pi/2}{1}) + C \Rightarrow 0 = 0 + C \Rightarrow C = 0$.So,$\ln y = \cos(x/y)$.For $y = 2$,we have $\ln 2 = \cos(x/2)$.We need to find $\cos(x(2))$. Since $\cos(x) = 2 \cos^2(x/2) - 1$,substituting $\cos(x/2) = \ln 2$ gives:$\cos(x(2)) = 2(\ln 2)^2 - 1$.

Let $x = x(y)$ be the solution of the differential equation $y = (x - y \frac{dx}{dy}) \sin(\frac{x}{y})$,$y > 0$ and $x(1) = \frac{\pi}{2}$. Then $\cos(x(2))$ is equal to:

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