2.1.26 \(x u_x+y u_y=u\) (Example 3.5.1 in Lokenath Debnath)

problem number 26

Added June 2, 2019.

From example 3.5.1, page 210 nonlinear pde’s by Lokenath Debnath, 3rd edition.

Solve for \(u(x,y)\)

\begin{align*} x u_x+y u_y&=u \end{align*}

Mathematica ✓

ClearAll["Global`*"]; 
pde = x*D[u[x, y], x] + y*D[u[x, y], y] ==u[x, y]; 
sol = AbsoluteTiming[TimeConstrained[DSolve[pde, u[x, y], {x, y}], 60*10]];
 

\[\left \{\left \{u(x,y)\to x c_1\left (\frac {y}{x}\right )\right \}\right \}\]

Maple ✓

restart; 
pde :=x*diff(u(x,y),x)+y*diff(u(x,y),y)=u(x,y); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,u(x,y))),output='realtime'));
 

\[u \left (x , y\right ) = f_{1} \left (\frac {y}{x}\right ) x\]

Hand solution

Solve

\[ xu_{x}+yu_{y}=u \]

Using the Lagrange-charpit method

\[ \frac {dx}{x}=\frac {dy}{y}=\frac {du}{u}\]

The first pair of equations gives

\[ x=C_{1}y \]

And \(\frac {dx}{x}=\frac {du}{u}\) gives

\[ x=C_{2}u \]

Since \(C_{2}=G\left ( C_{1}\right ) \) then \(\frac {x}{u}=G\left ( \frac {x}{y}\right ) \) or \(u=xG^{-1}\left ( \frac {x}{y}\right ) \). Let \(G^{-1}=F\). Then the solution

\[ u\left ( x,y\right ) =xF\left ( \frac {x}{y}\right ) \]

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