#### 2.1.28 $$x^2 u_x+y^2 u_y=(x+y) u$$ Example 3.5.3 in Lokenath Debnath

problem number 28

From example 3.5.3, page 211 nonlinear pde’s by Lokenath Debnath, 3rd edition.

Solve for $$u(x,y)$$ \begin {align*} x^2 u_x+y^2 u_y&=(x+y) u \end {align*}

Mathematica

$\left \{\left \{u(x,y)\to x y c_1\left (\frac {1}{x}-\frac {1}{y}\right )\right \}\right \}$

Maple

$u \left (x , y\right ) = x y \mathit {\_F1} \left (\frac {x -y}{x y}\right )$

Hand solution

Solve $x^{2}u_{x}+y^{2}u_{y}=\left ( x+y\right ) u$ Using the Lagrange-charpit method$\frac {dx}{x^{2}}=\frac {dy}{y^{2}}=\frac {du}{\left ( x+y\right ) u}$ The ﬁrst pair of equations gives \begin {equation} -\frac {1}{x}=-\frac {1}{y}+C_{1}\tag {1} \end {equation} And $$\frac {dx}{x^{2}}=\frac {du}{\left ( x+y\right ) u}$$ gives$\frac {\left ( x+y\right ) }{x^{2}}dx=\frac {du}{u}$ But from (1) $$\frac {1}{y}=\frac {1}{x}+C_{1}$$ or $$\frac {1}{y}=\frac {1+xC_{1}}{x}$$ or $$y=\frac {x}{1+xC_{1}}$$Therefore the above becomes\begin {align*} \frac {\left ( x+\frac {x}{1+xC_{1}}\right ) }{x^{2}}dx & =\frac {du}{u}\\ \frac {\left ( x\left ( 1+xC_{1}\right ) +x\right ) }{x^{2}\left ( 1+xC_{1}\right ) }dx & =\frac {du}{u}\\ \frac {2x+x^{2}C_{1}}{x^{2}+x^{3}C_{1}}dx & =\frac {du}{u}\\ \frac {2+xC_{1}}{x+x^{2}C_{1}}dx & =\frac {du}{u}\\ 2\ln x-\ln \left ( 1+C_{1}x\right ) & =\ln u+C_{2}\\ \ln \frac {x^{2}}{1+C_{1}x} & =\ln u+C_{2}\\ \frac {x^{2}}{1+C_{1}x} & =C_{2}u \end {align*}

But $$C_{1}=\frac {1}{y}-\frac {1}{x}$$ hence the above becomes\begin {align*} \frac {x^{2}}{1+\left ( \frac {1}{y}-\frac {1}{x}\right ) x} & =C_{2}u\\ \frac {x^{2}}{1+\left ( \frac {x-y}{yx}\right ) x} & =C_{2}u\\ \frac {yx^{2}}{y+\left ( x-y\right ) } & =C_{2}u\\ yx & =C_{2}u \end {align*}

Since $$C_{2}=G\left ( C_{1}\right )$$ then $$\frac {yx}{u}=G\left ( \frac {1}{y}-\frac {1}{x}\right )$$ or \begin {align*} u & =yxG^{-1}\left ( \frac {1}{y}-\frac {1}{x}\right ) \\ & =yxG^{-1}\left ( \frac {y-x}{yx}\right ) \end {align*}

Let $$G^{-1}=F$$ $u=yxF\left ( \frac {y-x}{yx}\right )$

Where $$F$$ is arbitrary function.

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