Determine an interval in which the solution to the given IVP exists.
$$(x-3){dy\over dx}+(\ln x)y=2x; (1,2)$$
I was wondering if I needed to to separation of variables or the thing where I multiply by $\mu$...
If someone could help point me in the right direction, that would be grand, thanks!
1 Answer
$\begingroup$We are given:
$$\tag 1 (x-3){dy\over dx}+(\ln x)y=2x; y(1) = 2$$
For a first-order linear DEQ, we consider the following IVP.
$$y'(x) + p(x)y(x) = g(x), y(x_0) = y_0.$$
If $p(x)$ and $g(x)$ are continuous functions on an open interval $\alpha \lt x \lt \beta$ and the interval contains $x_0$, then there is a unique solution to the IVP on that interval.
Some important notes:
- We may not be able to find a closed form solution, but this theorem tells us that it exists and there will be only one of them.
- If the interval in the theorem is the largest possible interval where $p(x)$ and $g(x)$ are continuous, then the interval is the interval of validity for the solution.
- We can find several intervals, but only one will contain $x_0$ for the problem and $y_0$ can be any value.
Lets rewrite $(1)$ in this form, yielding:
$$\dfrac{dy}{dx} + \dfrac{\ln x}{x-3}y(x) = \dfrac{2x}{x-3}, ~~ y(1) = 2.$$
For $p(x)$ and $g(x)$, we can see two issues.
- $(1)$ For $\ln x$, we need $x \gt 0$.
- $(2)$ For $x-3$, we need $x \ne 3$.
From these, we arrive at two intervals, $0 \lt x \lt 3$ and $3 \lt x \lt \infty$.
Finally, for this IVP, the first interval contains $x_0$ and is our interval of existence.
There is no closed form for this problem, but you can use Euler's or Runge-Kutta to numerically verify this.
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