Finding the general solution to $2u_x-3u_y+(U-x)=0$

Question

The PDE I'm working on is: $$2u_x-3u_y+(U-x)=0.$$

Using the method of characteristics I obtained $c_1=2x+3y.$ Where I am stuck is on $c_2$; currently I'm exploring $$\frac{dx}{2}=\frac{du}{u-x}.$$

My first instinct was to integrate as is and get the following: $$\frac{x}{2}+c_2=\log(u-x)$$ or $$x+c_2=2\log(u-x)$$ which becomes $$e^xe^{c_2}=(u-x)^2$$ Solving for $c_2$ we get $(u-x)^2e^{-x}=c_2$

Hence $F(2x+3y)=(u-x)^2e^{-x}.$

Am I on the right track for a general solution?

Answer 1

For the equation \begin{align} 2 u_{x} - 3 u_{y} + u - x = 0 \end{align} it is seen that \begin{align} \frac{dx}{2} = \frac{dy}{-3} = \frac{du}{x-u}. \end{align} Solving for the $x-y$ equation, $2 dy + 3 dx = 0$, leads to $c_{1} = 3 x + 2 y$. Consider the second characteristic from \begin{align} dy = \frac{3 \, du}{u-x}. \end{align} This leads to $c_{2} = y - 3 \ln(u-x)$, or \begin{align} u = x + e^{(y+c_{2})/3}. \end{align} Making appropriate connections of the characteristics $c_{1}$ and $c_{2}$, it can be seen that a solution is \begin{align} u(x,y) = e^{-(x + y/3)} + x -2. \end{align}

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Verification

From the solution proposed solution it can be seen that: \begin{align} u_{x} &= 1 - e^{-(x+y/3)} \\ u_{y} &= - \frac{1}{3} e^{-(x+y/3)} \\ \end{align} and leads to \begin{align} 2 u_{x} - 3 u_{y} &= 2 - e^{-(x+y/3)} \\ &= x - \left[ e^{-(x+y/3)} + x - 2 \right] \\ &= x-u \end{align} which becomes \begin{align} 2 u_{x} - 3 u_{y} + u = x. \end{align}

Answer 2

You made three mistakes:

1. The solution to $\frac{dx}{2}=-\frac{dy}{3}$ is not $2x+3y=c_1$, but $$ 3x+2y=c_1; \tag{1} $$

2. There is a sign error in your second equation; the correct equation to be solved is $$ \frac{dx}{2}=\frac{du}{x-u}; \tag{2} $$

3. Your solution (of the incorrect ODE) also is incorrect.

To solve $(2)$, let's rewrite it as $$ \frac{du}{dx}=\frac{x-u}{2} \implies \frac{du}{dx}+\frac{u}{2}=\frac{x}{2}. \tag{3} $$ Multiplying both sides of $(3)$ by $e^{x/2}$, we get $$ e^{x/2}\left(\frac{du}{dx}+\frac{u}{2}\right)=\frac{d}{dx}\left(e^{x/2}u\right)=\frac{x}{2}e^{x/2} $$ $$ \implies u =e^{-x/2}\int\frac{x}{2}e^{x/2}\,dx = x-2+c_2e^{-x/2}. \tag{4} $$ It follows from $(1)$ and $(4)$ that the general solution to the PDE is $$ u=x-2+e^{-x/2}f(3x+2y), \tag{5} $$ where $f$ is an arbitrary differentiable function.