2.1.11 \(3 u_x + 5 u_y = x\)

problem number 11

Taken from Mathematica help pages

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

\[ 3 u_x + 5 u_y = x \]

Mathematica ✓

ClearAll["Global`*"]; 
sol = AbsoluteTiming[TimeConstrained[DSolve[3*D[u[x, y], x] + 5*D[u[x, y], y] == x, u[x, y], {x, y}], 60*10]];
 

\[\left \{\left \{u(x,y)\to \frac {x^2}{6}+c_1\left (y-\frac {5 x}{3}\right )\right \}\right \}\]

Maple ✓

restart; 
interface(showassumed=0); 
pde :=3*diff(u(x, y), x) + 5*diff(u(x, y), y) = x; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,u(x,y))),output='realtime'));
 

\[u \left (x , y\right ) = \frac {x^{2}}{6}+f_{1} \left (-\frac {5 x}{3}+y \right )\]

Hand solution

Solve

\begin{align} 3u_{x}+5u_{y} & =x\nonumber \\ u_{x}+\frac {5}{3}u_{y} & =\frac {x}{3}\tag {1}\end{align}

Solution

Let \(u=u\left ( y\left ( x\right ) ,x\right ) \). Then

\begin{equation} \frac {du}{dx}=\frac {\partial u}{\partial y}\frac {dy}{dx}+\frac {\partial u}{\partial x}\tag {2}\end{equation}

Comparing (1),(2) shows that

\begin{align} \frac {du}{dx} & =\frac {x}{3}\tag {3}\\ \frac {dy}{dx} & =\frac {5}{3}\tag {4}\end{align}

Solving (3) gives

\begin{align*} u & =\frac {x^{2}}{6}+C_{1}\\ C_{1} & =u-\frac {x^{2}}{6}\end{align*}

From (4)

\begin{align*} y & =\frac {5}{3}x+C_{2}\\ C_{2} & =y-\frac {5}{3}x \end{align*}

Let \(C_{1}=F\left ( C_{2}\right ) \) where \(F\) is arbutrary function. This gives

\begin{align*} u-\frac {x^{2}}{6} & =F\left ( y-\frac {5}{3}x\right ) \\ u\left ( x,y\right ) & =F\left ( y-\frac {5}{3}x\right ) +\frac {x^{2}}{6}\end{align*}

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