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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?