Differential Equations Question 195
Question: The solution of the differential equation $ x^{2}\frac{dy}{dx}\cos \frac{1}{x}-y\sin \frac{1}{x}=-1 $ , where $ y\to -1asx\to \infty $ is
Options:
A) $ y=\sin \frac{1}{x}-\cos \frac{1}{x} $
B) $ y=\frac{x+1}{x\sin \frac{1}{x}} $
C) $ y=\cos \frac{1}{x}+sin\frac{1}{x} $
D) $ y=\frac{x+1}{x\cos 1/x} $
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Answer:
Correct Answer: A
Solution:
[a] $ x^{2}\frac{dy}{dx}\cos \frac{1}{x}-y\sin \frac{1}{x}=-1 $
Or $ \frac{dy}{dx}-\frac{y}{x^{2}}\tan \frac{1}{x}=-\sec \frac{1}{x}\frac{1}{x^{2}}(linear) $
I.F. $ ={e^{-\int{\frac{1}{x^{2}}\tan \frac{1}{x}dx}}}=\sec \frac{1}{x} $
Thus, solution is $ y\sec \frac{1}{x}=-\int{{{\sec }^{2}}( \frac{1}{x} )\frac{1}{x^{2}}dx=\tan \frac{1}{x}+c} $
Given $ y\to -1,x\to \infty .\text{Thus,}C=-1. $ Hence, equation of curve is $ y=\sin \frac{1}{x}-\cos \frac{1}{x} $ .
BETA
