2.1.11 $3u_x+5u_y=x$

problem number 11

Taken from Mathematica help pages

Solve for $u(x,y)$ $$3u_x+5u_y=x$$

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]];
 

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

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'));
 

$$u(x,y)=\frac{x^2}{6}+\mathit{\text{\_F1}}(-\frac{5x}{3}+y)$$

Hand solution

Solve$$\begin{array}{rl}3u_x+5u_y& =x\\ \text{(1)}& u_x+\frac{5}{3}{u}_y& =\frac{x}{3}\end{array}$$

Solution

Let $u=u(y\left(x\right),x)$. Then $$\begin{array}{}\text{(2)}& \frac{du}{dx}=\frac{\partial u}{\partial y}\frac{dy}{dx}+\frac{\partial u}{\partial x}\end{array}$$ Comparing (1),(2) shows that $$\begin{array}{}\text{(3)}& \frac{du}{dx}& =\frac{x}{3}\text{(4)}& \frac{dy}{dx}& =\frac{5}{3}\end{array}$$

Solving (3) gives$$\begin{array}{rl}u& =\frac{x^2}{6}+C_1\\ C_1& =u-\frac{x^2}{6}\end{array}$$

From (4)$$\begin{array}{rl}y& =\frac{5}{3}x+C_2\\ C_2& =y-\frac{5}{3}x\end{array}$$

Let $C_1=F\left(C_2\right)$ where $F$ is arbutrary function. This gives$$\begin{array}{rl}u-\frac{x^2}{6}& =F(y-\frac{5}{3}x)\\ u(x,y)& =F(y-\frac{5}{3}x)+\frac{x^2}{6}\end{array}$$

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