- #1
jjr
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Homework Statement
The problem is from Walter Gautschi - Numerical Analysis, exercise 5.1.
Consider the initial value problem
[itex]\frac{dy}{dx}=\kappa(y+y^3), 0\leq x\leq1; y(0)=s[/itex]
where [itex]\kappa > 0[/itex] (in fact, [itex]\kappa >> 1[/itex]) and s > 0. Under what conditions on s does the solution [itex]y(x) = y(x;s)[/itex] exist on the whole interval [0,1]? {Hint: find y explicitly.}
The Attempt at a Solution
Following the hint, I tried solving it as a separable differential equation:
[itex]\frac{dy}{y+y^3} = \kappa dx[/itex]
(Using wolframalpha here)
[itex]log(y) - \frac{1}{2}log(y^2+1) = \kappa x[/itex]
[itex]10^{log(y)-\frac{1}{2}log(y^2+1)} = 10^{\kappa x}[/itex]
Ending up with
[itex]\frac{y}{2} + \frac{1}{2y} = 10^{-\kappa x}[/itex]
Not sure how to solve this, so used wolframalpha again and got:
[itex]y = 10^{-\kappa x} \pm \sqrt{10^{-2 \kappa x} - 1} [/itex]
Evidently [itex]y(0) = 1[/itex], which means that s has to be equal to 1.
This doesn't seem right to me. Especially considering the work needed to find y explicitly, seeing as how this is not a course in differential equations. So what I am missing here?