Taken from Mathematica Symbolic PDE document
nonlinear first-order PDE, the Clairaut equation
Solve for \(u(x,y)\) \[ x u_x + y u_y + \frac {1}{2} ( (u_x)^2+ (u_y)^2 ) = 0 \]
Mathematica ✓
ClearAll["Global`*"]; pde = u[x, y] == x*D[u[x, y], {x}] + y*D[u[x, y], {y}] + (1/2)*(D[u[x, y], {x}]^2 + D[u[x, y], {y}]^2); sol = AbsoluteTiming[TimeConstrained[Simplify[DSolve[pde, u[x, y], {x, y}]], 60*10]];
\[\left \{\left \{u(x,y)\to c_1 x+c_2 y+\frac {1}{2} \left (c_1{}^2+c_2{}^2\right )\right \}\right \}\]
Maple ✓
restart; interface(showassumed=0); pde := x*diff(u(x, y), x) + y*diff(u(x, y),y) + 1/2 * ( diff(u(x, y), x)^2 + diff(u(x, y), y)^2)=0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,u(x,y),'build')),output='realtime'));
\[u \left (x , y\right ) = -\frac {x^{2}}{2}-\frac {y^{2}}{2}-\mathit {\_c}_{1} \ln \left (x +\sqrt {x^{2}+2 \mathit {\_c}_{1}}\right )+\mathit {\_c}_{1} \ln \left (y +\sqrt {y^{2}-2 \mathit {\_c}_{1}}\right )+c_{1}+c_{2}-\frac {\sqrt {x^{2}+2 \mathit {\_c}_{1}}\, x}{2}-\frac {\sqrt {y^{2}-2 \mathit {\_c}_{1}}\, y}{2}\]
Hand solution
Assuming the solution is \(u\left ( x,y\right ) =X\left ( x\right ) +Y\left ( y\right ) \). Substituting this into the PDE gives\begin {align*} xX^{\prime }+yY^{\prime }+\frac {1}{2}\left ( \left ( X^{\prime }\right ) ^{2}+\left ( Y^{\prime }\right ) ^{2}\right ) & =0\\ \frac {1}{2}\left ( X^{\prime }\right ) ^{2}+xX^{\prime } & =-\frac {1}{2}\left ( Y^{\prime }\right ) ^{2}-yY^{\prime } \end {align*}
The above is possible when each side is equal to same constant, say \(C_{1}\). This gives two ODE’s\begin {align} \frac {1}{2}\left ( X^{\prime }\right ) ^{2}+xX^{\prime } & =C_{1}\tag {1}\\ \frac {1}{2}\left ( Y^{\prime }\right ) ^{2}+yY^{\prime } & =-C_{1}\tag {2} \end {align}
ODE (1) becomes\begin {align*} \left ( X^{\prime }\right ) ^{2}+2xX^{\prime }-2C_{1} & =0\\ X^{\prime } & =\frac {-b}{2a}\pm \frac {1}{2a}\sqrt {b^{2}-4ac}\\ & =\frac {-2x}{2}\pm \frac {1}{2}\sqrt {4x^{2}+8C_{1}}\\ & =-x\pm \sqrt {x^{2}+2C_{1}} \end {align*}
For the case \(X^{\prime }=-x+\sqrt {x^{2}+2C_{1}}\), the solution is \begin {align*} X\left ( x\right ) & =\int -x+\sqrt {x^{2}+2C_{1}}dx+C_{2}\\ & =-\frac {x^{2}}{2}+\frac {x\sqrt {x^{2}+2C_{1}}}{2}+C_{1}\ln \left ( x+\sqrt {x^{2}+2C_{1}}\right ) +C_{2} \end {align*}
For the case \(X^{\prime }=-x-\sqrt {x^{2}+2C_{1}}\), the solution is\begin {align*} X\left ( x\right ) & =\int -x-\sqrt {x^{2}+2C_{1}}dx+C_{2}\\ & =-\frac {x^{2}}{2}-\frac {x\sqrt {x^{2}+2C_{1}}}{2}-C_{1}\ln \left ( x+\sqrt {x^{2}+2C_{1}}\right ) +C_{2} \end {align*}
Combining the above two solutions to one gives\begin {equation} X\left ( x\right ) =-\frac {x^{2}}{2}\pm \frac {x\sqrt {x^{2}+2C_{1}}}{2}\pm C_{1}\ln \left ( x+\sqrt {x^{2}+2C_{1}}\right ) +C_{2}\tag {3} \end {equation} ODE (2) becomes\begin {align*} \left ( Y^{\prime }\right ) ^{2}+2yY^{\prime }+2C_{1} & =0\\ Y^{\prime } & =\frac {-b}{2a}\pm \frac {1}{2a}\sqrt {b^{2}-4ac}\\ & =\frac {-2y}{2}\pm \frac {1}{2}\sqrt {4y^{2}-8C_{1}}\\ & =-y\pm \sqrt {y^{2}-2C_{1}} \end {align*}
For the case \(Y^{\prime }=-y+\sqrt {x^{2}-2C_{1}}\), the solution is \begin {align*} Y\left ( y\right ) & =\int -y+\sqrt {y^{2}-2C_{1}}dy+C_{2}\\ & =\frac {-y^{2}}{2}+\frac {y\sqrt {y^{2}-2C_{1}}}{2}-C_{1}\ln \left ( y+\sqrt {y^{2}-2C_{1}}\right ) +C_{3} \end {align*}
For the case \(Y^{\prime }=-y-\sqrt {y^{2}-2C_{1}}\), the solution is\begin {align*} Y\left ( y\right ) & =\int -y-\sqrt {y^{2}+2C_{1}}dy+C_{2}\\ & =-\frac {y^{2}}{2}-\frac {y\sqrt {x^{2}-2C_{1}}}{2}+C_{1}\ln \left ( y+\sqrt {y^{2}-2C_{1}}\right ) +C_{3} \end {align*}
Combining the above two solutions to one gives\begin {equation} Y\left ( x\right ) =-\frac {y^{2}}{2}\pm \frac {y\sqrt {y^{2}-2C_{1}}}{2}\pm C_{1}\ln \left ( y+\sqrt {y^{2}-2C_{1}}\right ) +C_{3}\tag {4} \end {equation} From (3,4) the final solution is\begin {align*} u\left ( x,y\right ) & =X\left ( x\right ) +Y\left ( x\right ) \\ & =\left ( -\frac {x^{2}}{2}\pm \frac {x\sqrt {x^{2}+2C_{1}}}{2}\pm C_{1}\ln \left ( x+\sqrt {x^{2}+2C_{1}}\right ) +C_{2}\right ) +\left ( -\frac {y^{2}}{2}\pm \frac {y\sqrt {y^{2}-2C_{1}}}{2}\pm C_{1}\ln \left ( y+\sqrt {y^{2}-2C_{1}}\right ) +C_{3}\right ) \\ & =-\frac {x^{2}}{2}\pm \frac {x}{2}\sqrt {x^{2}+2C_{1}}\pm C_{1}\ln \left ( x+\sqrt {x^{2}+2C_{1}}\right ) -\frac {y^{2}}{2}\pm \frac {y}{2}\sqrt {y^{2}-2C_{1}}\pm C_{1}\ln \left ( y+\sqrt {y^{2}-2C_{1}}\right ) +C_{4} \end {align*}
Where \(C_{4}=C_{2}+C_{3}\).
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