2.1.19 \( 2 u_x + 5 u_y = u^2(x,y) + 1\)

problem number 19

Taken from Mathematica help pages

General solution for a quasilinear first-order PDE

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

\[ 2 u_x + 5 u_y = u^2(x,y) + 1 \]

Mathematica ✓

ClearAll["Global`*"]; 
pde =  2*D[u[x, y], x] + 5*D[u[x, y], y] == u[x, y]^2 + 1; 
sol =  AbsoluteTiming[TimeConstrained[Simplify[DSolve[pde, u[x, y], {x, y}]], 60*10]];
 

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

Maple ✓

restart; 
interface(showassumed=0); 
pde := 2* diff(u(x, y), x) + 5*diff(u(x, y), y) = u(x, y)^2 + 1; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde, u(x, y))),output='realtime'));
 

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

Hand solution

Solve for \(u\left ( x,y\right ) \) in \(2u_{x}+5u_{y}=u^{2}+1.\) Using the Lagrange-charpit method, the characteristic equations are

\[ \frac {dx}{2}=\frac {dy}{5}=\frac {du}{u^{2}+1}\]

From the first pair of equation we obtain

\begin{align*} 5dx & =2dy\\ 5x & =2y+C_{1}\\ C_{1} & =5x-2y \end{align*}

Now we can pick the pair \(\frac {dy}{5}=\frac {du}{u^{2}+1}\) or \(\frac {dx}{2}=\frac {du}{u^{2}+1}\) to solve for \(u\). It does not matter which. Using

\[ \frac {dx}{2}=\frac {du}{u^{2}+1}\]

Integrating gives

\begin{align*} \frac {1}{2}x & =\arctan \left ( u\right ) +C_{2}\\ C_{2} & =\frac {1}{2}x-\arctan \left ( u\right ) \end{align*}

\(C_{1}\) and \(C_{2}\) are always related by \(C_{2}=F\left ( C_{1}\right ) \) where \(F\) is arbitrary function. Hence

\begin{align*} \frac {1}{2}x-\arctan \left ( u\right ) & =F\left ( 5x-2y\right ) \\ \arctan \left ( u\right ) & =\frac {1}{2}x-F\left ( 5x-2y\right ) \\ u\left ( x,y\right ) & =\tan \left ( \frac {1}{2}x-F\left ( 5x-2y\right ) \right ) \end{align*}

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