Solution to y'=y/(2ylny+y-x)
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Solution to y'=y/(2ylny+y-x)

[From: ] [author: ] [Date: 11-05-25] [Hit: ]
we have 1/y = dx/dy = x.Now,......
It's a first-order nonlinear differential equation. The final solution is supposed to be this (follow link): http://www2.wolframalpha.com/Calculate/MSP/MSP62019ffc3ia9ad5cbd8000055eag9h0e523i081?MSPStoreType=image/gif&s=11&w=163&h=36
but I need to know the steps, please
Thanks in advance

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y'=y/(2ylny+y-x)

Think about x as the function you are solving for; x = x(y). First, since y' = dy/dx, we have 1/y' = dx/dy = x'. Taking the reciprocal of both sides then gives

x' = dx/dy = (2ylny+y-x) /y

yx' = 2ylny + y - x

yx' + x = 2ylny + y

Now, the left handside is a perfect derivative:

yx' + x = d/dy( xy)

and so

xy = integral{ (2ylny + y) dy }
= y^2 ln(y) - y^2/2 + y^2/2 + C
= y^2 ln(y) + C

and solving for x gives

x = y ln(y) + C/y
1
keywords: to,039,Solution,ylny,Solution to y'=y/(2ylny+y-x)
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