Solving $xu_x+(x+y)u_y=1$ with Method of characteristics

Question

Can anybody help me in making two characteristic functions of the equation

$$xu_x+(x+y)u_y=1$$

I have found one of the Char. by equating $$\frac{dx}{x}=du$$

By which I get $$ln(x)=u$$

How to form second equation?

Edit:

Initial conditions is $u(1,y)=y$

Is this correct ? $$\frac{dx}{x}=\frac{dy}{x+y}$$ $$(x+y)dx=x dy$$ $$ xdx=xdy-ydx$$ $$\frac{xdx}{x^2}=\frac{xdy-ydx}{x^2}$$ $$\frac{dx}{x}=\frac{xdy-ydx}{x^2}$$ $$\frac{dx}{x}=d\left(\frac{y}{x}\right)$$

On integrating we will get another char.

Answer 1

You can define the two equations: $$ \frac{dx} {ds}= x \text{ and } \frac{dy}{ds} = x+y$$ or equivalently $\frac{dx}{x} = \frac{dy}{x+y}.$ Thus, we have: $$\frac{dy}{dx} = \frac{x+y}{x}\implies \frac{dy}{dx} = 1 + \frac{y}{x} .$$ The ODE is of the form $\frac{dy}{dx} = F\left(\frac y x\right)$, so we can set $z(x) = \frac{y}{x}\implies \frac{dy}{dx} = z(x) + x\cdot z'(x) \implies 1 + z(x) =z(x) + x\cdot z'(x)\implies z'(x) = \frac{1}{x}$.

Thus:

$ z(x) = \ln|x| +c\implies \ldots \implies y(x) = x\cdot \ln|x| + x\cdot c$

Can you take it from here?

Answer 2

I'm not exactly sure how you were introduced to this material so this may look a bit different, but we typically search for characteristics $(x(t),y(t))$ and set $z(t) = u(x(t),y(t))$. If we ensure that $$\dot x = x \,\,\,\,\,\, \text{ and } \,\,\,\,\,\, \dot y = x+y$$ then by the chain rule, we know $$\dot z = \dot x u_x + \dot y u_y = xu_x + (x+y)u_y = 1.$$ The characteristics need to originate from the curve where we have data, so we need $x(0) =1$ and $y(0) = y_0$ for some arbitrary $y_0 \in \mathbb R$. Since $u(1,y) = y$, this will give $z(0) = y(0) = y_0.$ So your equations are \begin{align*} &\dot x = x, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, x(0) = 1,\\ &\dot y = x+y, \,\,\,\,\,\, y(0) = y_0,\\ &\dot z = 1, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, z(0)= y_0. \end{align*} These give $$x(t) = e^t, \,\,\,\,\,\,\,\,\,\, z(t) = t + y_0$$ and $$\dot y - y = e^t \,\,\, \implies \,\,\, \frac d {dt}(e^{-t} y) = 1 \,\,\, \implies \,\,\,\, e^{-t}y - y_0 = t \,\,\,\, \implies \,\,\,\, y(t) = y_0 e^t + te^t. $$ Then if we can invert the equations (i.e., solve for $t, y_0$ in terms of $x,y$) then we will have the solution. We immediately see $t = \log x$. Then$$ y_0 x + x\log x =y \,\,\,\, \implies \,\,\,\,\, y_0 = \frac{y}{x} - \log x.$$ Then the solution is $$u(x,y) = z(t(x,y); y_0(x,y)) = \log x + \frac y x - \log x = \frac y x.$$