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The solution of the differential equation $\frac{dy}{dx}=\frac{6}{x^2}$; y(1) = 3 is: |
$y=9+\frac{6}{x}$ $y=9-\frac{6}{x}$ $y=-9+\frac{6}{x}$ $y=-3-\frac{6}{x}$ |
$y=9-\frac{6}{x}$ |
The correct answer is Option (2) - $y=9-\frac{6}{x}$ $\frac{dy}{dx}=\frac{6}{x^2}⇒\int dy=\frac{6}{x^2}dx$ $y=-\frac{6}{x}+C$ as $y(1)=3⇒3=-6+C$ $⇒C=9$ so $y=9-\frac{6}{x}$ |
