4
$\begingroup$

Question:

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: $y(e) = 2$

Problem:

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.

$\endgroup$
  • $\begingroup$ 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
  • 3
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.