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2014-10-05T05:20:11+05:30

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We need to use the solution methods of the first order differential equations for this. This particular case, is easier and solvable also this way. 

Slope=\frac{dy}{dx}=\frac{sin\ x}{x^2}-\frac{2y}{x}\\ \\\frac{dy}{dx}+ (\frac{2}{x})y=\frac{sin\ x}{x^2}\\ \\x^2\frac{dy}{dx}+ (2x)y=sin\ x\\ \\L H S\ is\ derivative\ of\ x^2y\ wrt\ x.\\ \\\frac{d(x^2y)}{dx}= sin\ x\\ \\x^2y=  \int\limits^{}_{} {Sin\ x} \, dx = - Cos\ x + C\\ \\y = \frac{C-Cos\ x}{x^2}\\ \\(\frac{ \pi }{2},1)\ \ is\ on\ the \ curve.\\ \\1=\frac{C-0}{ \pi^2/4 }\\ \\C = \frac{4}{ \pi^2}\\ \\The\ desired\ Curve\ is\ y = \frac{\frac{4}{ \pi^2 }-Cos\ x}{x^2},\ \ x \neq 0\\


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Nice solution!!
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thanx n u r welcom