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Find the Particular Solution of the Differential Equation x2dy=ydxx^{2} d y=y d x

Question 2

Multiple Choice

Find the particular solution of the differential equation x2dy=ydxx^{2} d y=y d x if y=1y=1 when x=1x=1 .


A) 1y=1x+2\frac{1}{y}=-\frac{1}{x}+2 or y=x2x1y=\frac{x}{2 x-1}
B) lny=1x+1\ln y=-\frac{1}{x}+1 or y=e1x+1y=e^{\frac{-1}{x}+1}
C) lny=lnx2\ln y=\ln x^{2} or y=x2y=x^{2}
D) lny=1x+2\ln y=-\frac{1}{x}+2 or y=e1x+2y=e^{\frac{-1}{x}+2}

Correct Answer:

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