Solving : $2xy \cdot u_x + u_y - u = 0$

Question

Exercise :

Solve the Initial Value Problem (IVP) : $$2xy \cdot u_x + u_y - u = 0$$ $$u(x,0) = x$$

Attempt :

The functions $a(x,y) = 2xy$, $c(x,y) = 1$, $f(x,y) = 0$ are $C^1$ in an open set including the random point $(x_0,0)$ of the $x-$axis, while the function $\phi(x) = x$ is also $C^1$ in an open interval including the point $x_0$. Thus, via a known lemma, there exists a unique solution for the IVP in a neighborhood of $(x_0,0)$.

Yielding the Lagrange problem :

$$\frac{\mathrm{d}x}{2xy} = \frac{\mathrm{d}y}{1} = \frac{\mathrm{d}u}{u}$$

Question : Moving from here on, how would one yield the results that Wolfram Alpha calculates here ?

Answer 1

You have that $\frac{dx}{x} = 2ydy$. Integrating you get $\ln(x) = y^2 + C$

By integrating $\frac{du}{u} = dy$ we get $\ln(u) = y + f(C)$ and so $\ln(u) = y + f(\ln(x) - y^2)$ is your general solution for some real-valued differentiable function $f$.

To find the particular solution evaluate at $(x,0)$ to get:

$$\ln(x) = f(\ln(x)) \implies f(t) = t$$

Finally the wanted solution is:

$$\ln(u) = y - y^2 + \ln(x) \implies \boxed{u(x,y) = xe^{y-y^2}}$$

---

To get the general solution that Wolfram Alpha gives just exponentiate both sides and the set $F = e^f$ in my solution. Also the factor of $\frac 12$ is irrelevant as we take $f$ to be any real-valued differentiable function $f$.

I'm not sure why Wolfram takes $\pm$ in front of $y$. I mean the $-$ sign doesn't give you a solution. Take $F \equiv 1$ then $u=e^{-y}$, assuming $y$ is positive. However then we have:

$$2xyu_x + u_y - u = -e^{-y} - e^{-y} = -2e^{-y} \not = 0$$

Answer 2

An alternate solution, without use of WA:

From $$\frac{\mathrm{d}x}{2xy} = \frac{\mathrm{d}y}{1} = \frac{\mathrm{d}u}{u}$$ it can be seen that $$\frac{dx}{2 x y} = \frac{dy}{1} \hspace{5mm} \to \hspace{5mm} \frac{dx}{x} = 2 y \, dy$$ leads to $\ln x - y^{2} = c_{1}$ and $$ \frac{du}{u} = \frac{dy}{1}$$ leads to $\ln u - y = c_{2}$. A solution of the form $ u = e^{y + c_{2}} = e^{y} \, e^{f(\ln x - y^{2})}$ can be obtained. Now using $u(x,0) = x$ provides $$ x = e^{0} \, e^{f(\ln x - 0^{2})} = e^{f(\ln x)}$$ or $f(t) = t$ and leads to the solution $$u(x,y) = x \, e^{y - y^{2}}. $$

As a check of the solution: \begin{align} u_{x} &= e^{y - y^{2}} = \frac{u}{x} \\ u_{y} &= x \, (1 - 2 y) \, e^{y - y^{2}} = (1 -2 y) \, u \end{align} then $2 x y \, u_{x} + u_{y} = 2 y u + (1 - 2 y ) u = u$ as required.

A minor extension can be seen as: $$u = a x y \, u_{x} + b \, u_{y} \, \hspace{5mm} u(x,0) = x$$ has the solution $$u(x,y) = x \, e^{-\frac{a}{2 b} \, (y^2 - 2 y/a)}$$

Answer 3

$$2xyu_x+u_y=u$$ $\frac{dx}{2xy}=\frac{dy}{1}=\frac{du}{u}$

First characteristics, from $\quad\frac{dx}{2xy}=\frac{dy}{1}$ : $$xe^{-y^2}=c_1$$ Second characteristics, from $\quad \frac{dy}{1}=\frac{du}{u}$ : $$ue^{-y}=c_2$$ General solution : $\quad ue^{-y}=F\left(xe^{-y^2}\right)$ $$u(x,y)=e^yF\left(xe^{-y^2}\right)$$ $F$ is an arbitrary function, to be determined according to the boundary condition. $$ $$ Condition : $\quad u(x,0)=x=e^0F\left(xe^{-0^2}\right)=F(x)$

Now the function is determined : $\quad F(X)=X\quad$ any $X$.

We put it into the above general solution where $X=xe^{-y^2}\quad$ thus $\quad F\left(xe^{-y^2}\right) =xe^{-y^2}$.

The particular solution fitting to the boundary condition is : $\quad u(x,y)=e^y\left(xe^{-y^2}\right)$ $$u(x,y)=xe^{y-y^2}$$