#### 2.1.11 $$3 u_x + 5 u_y = x$$

problem number 11

Taken from Mathematica help pages

Solve for $$u(x,y)$$ $3 u_x + 5 u_y = x$

Mathematica

$\left \{\left \{u(x,y)\to \frac {x^2}{6}+c_1\left (y-\frac {5 x}{3}\right )\right \}\right \}$

Maple

$u \left (x , y\right ) = \frac {x^{2}}{6}+\mathit {\_F1} \left (-\frac {5 x}{3}+y \right )$

Hand solution

Solve\begin {align} 3u_{x}+5u_{y} & =x\nonumber \\ u_{x}+\frac {5}{3}u_{y} & =\frac {x}{3}\tag {1} \end {align}

Solution

Let $$u=u\left ( y\left ( x\right ) ,x\right )$$. Then \begin {equation} \frac {du}{dx}=\frac {\partial u}{\partial y}\frac {dy}{dx}+\frac {\partial u}{\partial x}\tag {2} \end {equation} Comparing (1),(2) shows that \begin {align} \frac {du}{dx} & =\frac {x}{3}\tag {3}\\ \frac {dy}{dx} & =\frac {5}{3}\tag {4} \end {align}

Solving (3) gives\begin {align*} u & =\frac {x^{2}}{6}+C_{1}\\ C_{1} & =u-\frac {x^{2}}{6} \end {align*}

From (4)\begin {align*} y & =\frac {5}{3}x+C_{2}\\ C_{2} & =y-\frac {5}{3}x \end {align*}

Let $$C_{1}=F\left ( C_{2}\right )$$ where $$F$$ is arbutrary function. This gives\begin {align*} u-\frac {x^{2}}{6} & =F\left ( y-\frac {5}{3}x\right ) \\ u\left ( x,y\right ) & =F\left ( y-\frac {5}{3}x\right ) +\frac {x^{2}}{6} \end {align*}

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