# How to find a specific curve if the initial value is not given?

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.

• y(e) = C....... – William Elliot Dec 30 '18 at 11:46
• @WilliamElliot Nah. Edited the question. – harshit54 Dec 30 '18 at 11:48
• Note that, evaluating the original equation with $x=1$, you find $y(1)=0$ – Martín Vacas Vignolo Dec 30 '18 at 11:59
• @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. – harshit54 Dec 30 '18 at 12:06
• 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 – Martín Vacas Vignolo Dec 30 '18 at 12:27

## 1 Answer

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 $$$$\lim_{x \searrow 1} y(x) = \left\{ \begin{array}{ll} -\infty, C < 2\\ 0, C = 2\\ \infty, C > 2 \end{array} \right..$$$$ Therefore, if we assume that the right-sided limit of $$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.