What is the issue with the two methods used to solve the ODE dy/dx=x(1-y)?

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The discussion revolves around solving the ordinary differential equation dy/dx = x(1-y) using both linear and separable methods, which yield different forms of the solution. The integrating factor method leads to the solution y = 1 + c*e^(-x^2/2), while the separable method results in y = 1 - 1/K*e^(x^2/2). Participants note that despite the differing appearances, both solutions can be equivalent under certain conditions, specifically with appropriate substitutions for constants. A key point raised is the importance of careful manipulation and consistency in constants when transitioning between forms. Ultimately, the two methods are valid and yield equivalent solutions when correctly interpreted.
badtwistoffate
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Got the eqn dy/dx=x(1-y) and it can be solved both linear and separable methods.(Linear method being using a integrating factor) Problem I am having is that with this two methods i get two different (yet similar answers) and was wondering if you can see my problem with this two methods I am using.

Integrating Factor method:
y'+xy=x, u'(x)=e^(x^2/2)

[e^(x^2/2)y]'=x*e^(x^2/2)

e^(x^2/2)y=integral(x*e^(x^2/2)), do u substitution, get...

e^(x^2/2)y=e^(x^2/2)+c

y=1+c/e^(x^2/2) or y=1+c*e^(-x^2/2)

Separable method:
dy/(1-y)=x dx, integrate both sides

-ln(1-y)=e^(x^2/2)+C, raise both sides to e.

1/(1-y)=K*e^(x^2/2)+C, rearrange to get y=.

y=1-1/K*e^(x^2/2)

so we get two different answers with these methods, where is the problem lieing or are both wrong?
 
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badtwistoffate said:
Separable method:
dy/(1-y)=x dx, integrate both sides

-ln(1-y)=e^(x^2/2)+C, raise both sides to e.

1/(1-y)=K*e^(x^2/2)+C, rearrange to get y=.

y=1-1/K*e^(x^2/2)
You got a bit sloppy near the end. Some mistakes are just typo's I think.

-\ln(1-y)=\frac{1}{2}x^2+C
1-y=K\exp(-\frac{1}{2}x^2)
y=1-K\exp(-\frac{1}{2}x^2)

So it's the same.
 
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