yux − xuy = eu with u(0,y) = y2 − 1

Mathematica ✓

ClearAll["Global`*"]; 
pde = y*D[u[x, y], x] - x*D[u[x, y], y] == Exp[u[x, y]]; 
sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, u[0, y] == y^2 - 1}, u[x, y], {x, y}], 60*10]];

\[\left \{\left \{u(x,y)\to -\log \left (e^{-x^2-y^2+1}+\tan ^{-1}\left (\frac {x}{\sqrt {y^2}}\right )\right )\right \}\right \}\]

Maple ✓

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

\[u \left ( x,y \right ) =\ln \left ( - \left ( \arctan \left ( {\frac {x}{y}} \right ) - \left ( {{\rm e}^{{x}^{2}+{y}^{2}-1}} \right ) ^{-1} \right ) ^{-1} \right ) \]

Solve \begin {align*} yu_{x}-xu_{y} & =e^{u}\\ u\left ( 0,y\right ) & =y^{2}-1 \end {align*}

The characteristic equations (using \(s\) as the parameter) are\begin {align} \frac {dx}{ds} & =y\tag {1}\\ \frac {dy}{ds} & =-x\tag {2}\\ \frac {du}{ds} & =e^{u} \tag {3} \end {align}

With initial point at \(s=0\) which can be written as \(u\left ( 0,\xi \right ) =\xi ^{2}-1\). The idea of this method is to use (1,2) to obtain expressions for \(s,\xi \). These are the unknowns. Then using these in (3) to obtain the ﬁnal solution \(u\left ( x,y\right ) \). One problem that we see right away, is that (1,2) are coupled. When this happens, we must decouple them ﬁrst. Diﬀerentiating (1) gives \(\frac {d^{2}x}{ds^{2}}=y^{\prime }=-x\). Hence (1) becomes \(x^{\prime \prime }\left ( s\right ) +x=0\). This has solution \begin {equation} x\left ( s\right ) =c_{1}\cos s+c_{2}\sin s \tag {4} \end {equation} But from (1), we see that \(x^{\prime }\left ( s\right ) =y\). Therefore \begin {equation} y\left ( s\right ) =-c_{1}\sin s+c_{2}\cos s \tag {5} \end {equation} We made some progress. Found \(x\left ( s\right ) ,y\left ( s\right ) \). But need to solve for \(c_{1},c_{2}\). This is done using the initial point, which is always at \(s=0\). When \(s=0\), using \(u\left ( 0,\xi \right ) =\xi ^{2}-1\), which says \(x\left ( 0\right ) =0\) and \(y\left ( 0\right ) =\xi \). Using \(x\left ( 0\right ) =0\) when \(s=0\) then from (4) we obtain \(0=c_{1}\). Now the solutions becomes\begin {align*} x\left ( s\right ) & =c_{2}\sin s\\ y\left ( s\right ) & =c_{2}\cos s \end {align*}

Now we use the second initial condition on \(y\left ( s\right ) \) which says \(y\left ( 0\right ) =\xi \). Hence from the second equation above, \(\xi =c_{2}\). Therefore the solution now becomes\begin {align} x\left ( s\right ) & =\xi \sin s\tag {6}\\ y\left ( s\right ) & =\xi \cos s \tag {7} \end {align}

This is as far as we can go. Remembering that we are after expressions for \(s\) and \(\xi \). Dividing (6/7) gives \begin {align} \frac {x}{y} & =\tan \left ( s\right ) \nonumber \\ s & =\arctan \left ( \frac {x}{y}\right ) \tag {8} \end {align}

Good. We obtained relation for \(s\) in terms of \(x,y\). What about \(\xi \)?. By equation (6) and (7) and adding them we obtain\begin {align} x^{2}\left ( s\right ) +y^{2}\left ( s\right ) & =\xi ^{2}\nonumber \\ \xi & =\sqrt {x^{2}+y^{2}} \tag {9} \end {align}

Good. (8,9) is what we wanted. Equations (1,2) have done their job. We used them to ﬁnd \(s,\xi \). Now we move on to (3) which is \[ \frac {du}{ds}=e^{u}\] Solving it gives\[ -e^{-u}=s+C \] But at \(s=0\) we know that \(u\left ( 0\right ) =\xi ^{2}-1\). \(-e^{-\left ( \xi ^{2}-1\right ) }=C\). Then the above becomes\[ -e^{-u}=s-e^{-\left ( \xi ^{2}-1\right ) }\] We are almost there. We just need now to go back to \(x,y\) from \(s,\xi \). By using (8,9) the above becomes\begin {align*} -e^{-u} & =\arctan \left ( \frac {x}{y}\right ) -e^{-\left ( x^{2}+y^{2}-1\right ) }\\ e^{-u} & =-\arctan \left ( \frac {x}{y}\right ) +e^{-\left ( x^{2}+y^{2}-1\right ) } \end {align*}

We can stop here. But if assume \(u>0\) then the above can be simpliﬁed more \begin {align*} -u & =\ln \left ( -\arctan \left ( \frac {x}{y}\right ) +e^{-\left ( x^{2}+y^{2}-1\right ) }\right ) \\ u\left ( x,y\right ) & =-\ln \left ( -\arctan \left ( \frac {x}{y}\right ) +e^{-\left ( x^{2}+y^{2}-1\right ) }\right ) \\ & =-\ln \left ( -\left ( \arctan \left ( \frac {x}{y}\right ) -e^{-\left ( x^{2}+y^{2}-1\right ) }\right ) \right ) \\ & =\ln \left ( \frac {1}{-\left ( \arctan \left ( \frac {x}{y}\right ) -e^{-\left ( x^{2}+y^{2}-1\right ) }\right ) }\right ) \end {align*}

I used document titled "The method of Characteristics" by Ryan C. Daileda for help which is a very useful document.

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2.1.70 \(y u_x - x u_y = e^u\) with \(u(0,y)=y^2-1\)