#### 6.1.5 problem number 5

problem number 420

Problem 1.5 from Handbook of ﬁrst order partial diﬀerential equations by Polyanin, Zaitsev, Moussiaux.

Solve for $$w(x,y)$$ $w_x = w f(x,y)+ g(x,y)$

Mathematica

$\left \{\left \{w(x,y)\to \exp \left (\int _1^xf(K[1],y)dK[1]\right ) \left (\int _1^x\exp \left (-\int _1^{K[2]}f(K[1],y)dK[1]\right ) g(K[2],y)dK[2]+c_1(y)\right )\right \}\right \}$

Maple

$w \left (x , y\right ) = \left (\int {\mathrm e}^{-\left (\int f \left (x , y\right )d x \right )} g \left (x , y\right )d x +\mathit {\_F1} (y )\right ) {\mathrm e}^{\int f \left (x , y\right )d x}$

Hand solution

\begin {align*} \frac {\partial w}{\partial x} & =wf\left ( x,y\right ) +g\left ( x,y\right ) \\ \frac {\partial w}{\partial x}-wf\left ( x,y\right ) & =g\left ( x,y\right ) \end {align*}

We can treat this similar to linear ODE and use an integrating factor $$I=e^{-\int f\left ( x,y\right ) dx}$$ hence the above becomes\begin {align*} \frac {\partial }{\partial x}\left ( we^{-\int f\left ( x,y\right ) dx}\right ) & =e^{-\int wdx}g\left ( x,y\right ) \\ we^{-\int f\left ( x,y\right ) dx} & =\int e^{-\int f\left ( x,y\right ) dx}g\left ( x,y\right ) dx+G\left ( y\right ) \\ w & =\left ( \int e^{-\int f\left ( x,y\right ) dx}g\left ( x,y\right ) dx+G\left ( y\right ) \right ) e^{\int f\left ( x,y\right ) dx} \end {align*}

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