#### 2.1.71 $$y u_x - x u_y = e^u$$

problem number 71

Added May 21, 2019.

This is same problem as above, but without I.C. given.

Solve for $$u(x,y)$$ \begin {align*} y u_x - x u_y = e^u \end {align*}

No IC are given.

Mathematica

$\left \{\left \{u(x,y)\to -\log \left (\tan ^{-1}\left (\frac {x}{\sqrt {y^2}}\right )-c_1\left (\frac {1}{2} \left (x^2+y^2\right )\right )\right )\right \}\right \}$

Maple

$u \left (x , y\right ) = \ln \left (-\frac {1}{\mathit {\_F1} \left (x^{2}+y^{2}\right )+\arctan \left (\frac {x}{y}\right )}\right )$

Hand solution

Solve $yu_{x}-xu_{y}=e^{u}$

Since no initial conditions are given, I ﬁnd using Lagrange-charpit method is better here than using characteristic equations with parameter $$s$$. The Lagrange-charpit equations for the above PDE are

$\frac {dx}{y}=\frac {-dy}{x}=\frac {du}{e^{u}}$

The ﬁrst two equations gives

\begin {align} xdx & =-ydy\nonumber \\ \frac {x^{2}}{2} & =-\frac {y^{2}}{2}+C_{1}\nonumber \\ C_{1} & =\frac {1}{2}\left ( x^{2}+y^{2}\right ) \nonumber \\ C_{1} & =\left ( x^{2}+y^{2}\right ) \tag {1} \end {align}

Where the $$2$$ is absorbed by the constant. We now need to decide to either solve $$\frac {-dy}{x}=\frac {du}{e^{u}}$$ together or $$\frac {dx}{y}=\frac {du}{e^{u}}$$. It does not matter which pair to pick. Using the second pair gives

$\frac {dx}{y}=\frac {du}{e^{u}}$

But from (1), $$y=\sqrt {C_{1}-x^{2}}$$ (taking only the positive root) and the above equation now becomes

$\frac {dx}{\sqrt {C_{1}-x^{2}}}=\frac {du}{e^{u}}$

Integrating gives

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

In this method, the constants $$C_{1},C_{2}$$ are always related by $$C_{2}=F\left ( C_{1}\right )$$ where $$F$$ is an arbitrary function. Hence we obtain

\begin {align*} \arctan \left ( \frac {x}{y}\right ) +e^{-u} & =F\left ( x^{2}+y^{2}\right ) \\ e^{-u} & =F\left ( x^{2}+y^{2}\right ) -\arctan \left ( \frac {x}{y}\right ) \end {align*}

For positive $$u$$ the above simpliﬁes to

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

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