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nikdorinn [45]
3 years ago
13

Im confused on this one !

Mathematics
1 answer:
ludmilkaskok [199]3 years ago
5 0

I think that since the x values, as they increase become farther and farther from the x-axis, the correct point is (1.5, -3). Because on both sides of the y-axis they are 1.5 units away, since one is positive and one is negative.


Hope this helps! :)

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The general solution of 2 y ln(x)y' = (y^2 + 4)/x is
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Integrate both sides; on the left, set u=y^2+4 so that \mathrm du=2y\,\mathrm dy; on the right, set v=\ln x so that \mathrm dv=\dfrac{\mathrm dx}x. Then

\displaystyle\int\frac{2y}{y^2+4}\,\mathrm dy=\int\dfrac{\mathrm dx}{x\ln x}\iff\int\frac{\mathrm du}u=\int\dfrac{\mathrm dv}v

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