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
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
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