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## Solution to No. 1

Solution to No. 1$(x - y \ln y + y \ln x) \, dx + x(\ln y - \ln x) \, dy = 0$

$\big[ x - y(\ln y - \ln x) \big] \, dx + x(\ln y - \ln x) \, dy = 0$

$\left[ x - y\ln \left( \dfrac{y}{x} \right) \right] dx + x\ln \left( \dfrac{y}{x} \right) dy = 0$

y = vx

dy = v dx + x dv

$\left[ x - vx\ln \left( \dfrac{vx}{x} \right) \right] dx + \left[ x\ln \left( \dfrac{vx}{x} \right) \right](v \, dx + x \, dv) = 0$

$(x - vx\ln v) \, dx + (x\ln v)(v \, dx + x \, dv) = 0$

$x \, dx - vx\ln v \, dx + vx\ln v \, dx + x^2 \ln v \, dv = 0$

$x \, dx + x^2 \ln v \, dv = 0$

$\dfrac{dx}{x} + \ln v \, dv = 0$

$\ln x + (v \ln v - v) = c$

$\ln x + \dfrac{y}{x} \ln \left( \dfrac{y}{x} \right) - \dfrac{y}{x} = c$

## pano po ung problem 2,3,4 for

In reply to Solution to No. 1 by Jhun Vert

pano po ung problem 2,3,4 for homogeneous DE

## how about the problem no.2,3

how about the problem no.2,3 and4

## naka graduate kana sir?

In reply to how about the problem no.2,3 by Sydney Sales

naka graduate kana sir?