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


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

Maple


\[u \left (x , y\right ) = \tan \left (\frac {x}{2}+\frac {\mathit {\_F1} \left (-\frac {5 x}{2}+y \right )}{2}\right )\]

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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