2.1.74 \(2 u_x + 3 u_y = 1\)

problem number 74

Added May 23, 2019.

From Math 5587 midterm I, Fall 2016, practice exam, problem 5.

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

\begin{align*} 2 u_x + 3 u_y = 1 \end{align*}

Mathematica ✓

ClearAll["Global`*"]; 
pde = 2 D[u[x, y], x] + 3*D[u[x, y], y] == 1; 
sol = AbsoluteTiming[TimeConstrained[DSolve[pde, u[x, y], {x, y}], 60*10]];
 

\[\left \{\left \{u(x,y)\to \frac {x}{2}+c_1\left (y-\frac {3 x}{2}\right )\right \}\right \}\]

Maple ✓

restart; 
pde :=2*diff(u(x,y),x)+3*diff(u(x,y),y)=1; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,u(x,y))),output='realtime'));
 

\[u \left (x , y\right ) = \frac {x}{2}+f_{1} \left (-\frac {3 x}{2}+y \right )\]

Hand solution

Solve \(2u_{x}+3u_{y}=1\). Using the method of characteristics

\[ \frac {dx}{2}=\frac {dy}{3}=\frac {du}{1}\]

From the first pair of equations we obtain \(\frac {1}{2}x=\frac {1}{3}y+C_{1}\) or \(C_{1}=\frac {1}{2}x-\frac {1}{3}y\). From the pair \(\frac {dx}{2}=\frac {du}{1}\) we obtain

\(\frac {1}{2}x=u+C_{2}\) or \(C_{2}=\frac {1}{2}x-u\). But \(C_{2}=F\left ( C_{1}\right ) \) where \(F\) is arbitrary function. Hence

\begin{align*} \frac {1}{2}x-u & =F\left ( \frac {1}{2}x-\frac {1}{3}y\right ) \\ u & =\frac {1}{2}x-F\left ( \frac {1}{2}x-\frac {1}{3}y\right ) \end{align*}

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