Let $y(x)$ be the solution of the differential equation

$x\cdot ln(x)\dfrac{dy}{dx}+y=2x\cdot ln(x)$, $x\ge1$.

Find $y(e)$.

Answer: $y(e) = 2$


So I understand that this can be converted into a simple linear differential equation and found that the solution is:

$y\cdot ln(x)=2(x\cdot ln(x) - x) + C$

This is a family of curves. However for solving the question, I need a specific curve out of all these. What I don't understand is how how do I find that particular curve as the initial value of the function is not given.

  • $\begingroup$ y(e) = C....... $\endgroup$ – William Elliot Dec 30 '18 at 11:46
  • $\begingroup$ @WilliamElliot Nah. Edited the question. $\endgroup$ – harshit54 Dec 30 '18 at 11:48
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    $\begingroup$ Note that, evaluating the original equation with $x=1$, you find $y(1)=0$ $\endgroup$ – Martín Vacas Vignolo Dec 30 '18 at 11:59
  • $\begingroup$ @MartínVacasVignolo So there is only one possible solution for this equation. But how is it possible if I am not given the initial values for the DE. Because the whole family satisfies the given DE. $\endgroup$ – harshit54 Dec 30 '18 at 12:06
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    $\begingroup$ hmm no, without conditions you must consider the maximal domain. In this case $\mathbb{R}^+$, and because $1\in\mathbb{R}^+$, is an equivalent problem $\endgroup$ – Martín Vacas Vignolo Dec 30 '18 at 12:27

I'd say that the answer $y(e) = 2$ is wrong if the question is stated like this.

You have already found the correct general solution of the first-order linear ODE, with one constant of integration $C \in \mathbb{R}$. From this you do indeed get $y(e) = C$ as William Elliot has pointed out. Therefore, you can obtain any value for $y(e)$, depending on the value of $C$.

The reason why the particular solution with $C=2$ is of interest is that it is the only solution of this ODE for which the (right-sided) limit as $x \searrow 1$ is finite.

For the general solution $y(x) = \frac{2x(\ln(x)-1)+C}{\ln(x)}$, $C \in \mathbb{R}$, we have \begin{equation} \lim_{x \searrow 1} y(x) = \left\{ \begin{array}{ll} -\infty, C < 2\\ 0, C = 2\\ \infty, C > 2 \end{array} \right.. \end{equation} Therefore, if we assume that the right-sided limit of $y(x)$ is finite as $x$ approaches $1$, then there remains only the particular solution with $C=2$, which satisfies $y(e) = 2$.

But this assumption needs to be added to the question, otherwise the answer is wrong.


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