- #1
kingwinner
- 1,270
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1) Solve the initial value problem (x - xy) + (y + x2) dy/dx = 0, y(0)=2
Hint: The ODE is not exact, but can be made exact. There exists an integrating factor of the form u=u(x2+y2)
M(x,y) + N(x,y) y' = 0
Omitting the hint, I found out that
(Nx - My) / 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!
Hint: The ODE is not exact, but can be made exact. There exists an integrating factor of the form u=u(x2+y2)
M(x,y) + N(x,y) y' = 0
Omitting the hint, I found out that
(Nx - My) / 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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