- #1

jjr

- 51

- 1

## 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?