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**1) Solve the initial value problem (x - xy) + (y + x**

Hint: The ODE is not exact, but can be made exact. There exists an integrating factor of the form u=u(x

^{2}) dy/dx = 0, y(0)=2Hint: The ODE is not exact, but can be made exact. There exists an integrating factor of the form u=u(x

^{2}+y^{2})M(x,y) + N(x,y) y' = 0

Omitting the hint, I found out that

(N

_{x}- M

_{y}) / M = 3 / (1-y) which is just a function of y

So u(y) = exp∫ 3 / (1-y) dy = 1 / |1-y|

^{3}must be an integrating factor which can make the given ODE exact.

But how can I cope with the absolute value? I can't integrate until I can get rid of it. Now, if I just take

**one**of u(y) = 1 / (1-y) or u(y) = 1 / (y-1), is that OK? Will I be getting the same general solution?

Another thing is that the initial condition is x=0, y=2>1, so

**must**we take u(y) = 1 / (y-1) and not u(y) = 1 / (1-y)?

Secondly, I am very interested in what integrating factor the hint is referring to, what does it mean and

**how**can I find it?

Can someone help? I would really appreciate!

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