The solution of the differential equation,x^2 | vedclass

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(A) Given the differential equation: $x^2 \frac{dy}{dx} \cos \frac{1}{x} - y \sin \frac{1}{x} = -1$. Dividing by $x^2 \cos \frac{1}{x}$,we get: $\frac

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$y = \sin \frac{1}{x} - \cos \frac{1}{x}$

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(A) Given the differential equation: $x^2 \frac{dy}{dx} \cos \frac{1}{x} - y \sin \frac{1}{x} = -1$. Dividing by $x^2 \cos \frac{1}{x}$,we get: $\frac{dy}{dx} - y \frac{	an(1/x)}{x^2} = -\frac{\sec(1/x)}{x^2}$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -\frac{	an(1/x)}{x^2}$ and $Q = -\frac{\sec(1/x)}{x^2}$. The Integrating Factor $(IF)$ is $e^{\int P dx} = e^{\int -\frac{	an(1/x)}{x^2} dx}$. Let $u = \frac{1}{x}$,then $du = -\frac{1}{x^2} dx$. $IF = e^{\int 	an u du} = e^{\ln |\sec u|} = \sec(\frac{1}{x})$. The solution is $y \cdot IF = \int Q \cdot IF dx + c$. $y \sec(\frac{1}{x}) = \int -\frac{\sec(1/x)}{x^2} \cdot \sec(\frac{1}{x}) dx = -\int \sec^2(\frac{1}{x}) \cdot \frac{1}{x^2} dx$. Let $u = \frac{1}{x}$,then $du = -\frac{1}{x^2} dx$. $y \sec(\frac{1}{x}) = \int \sec^2 u du = 	an u + c = 	an(\frac{1}{x}) + c$. As $x ightarrow \infty$,$\frac{1}{x} ightarrow 0$. Given $y ightarrow -1$,we have: $-1 \cdot \sec(0) = 	an(0) + c \Rightarrow -1 = 0 + c \Rightarrow c = -1$. Thus,$y \sec(\frac{1}{x}) = 	an(\frac{1}{x}) - 1$. $y = \frac{	an(1/x) - 1}{\sec(1/x)} = \sin(\frac{1}{x}) - \cos(\frac{1}{x})$.

The solution of the differential equation,$x^2 \frac{dy}{dx} \cos \frac{1}{x} - y \sin \frac{1}{x} = -1,$ where $y \rightarrow -1$ as $x \rightarrow \infty$ is

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