Lagrange’s first-order linear partial DE Show that the general solution of the partial DE
$$u_{x} + (a(x)y + b(x)u)u_{y}=c(x)y + d(x)u  $$
is of the form 
$$\phi \left ( \frac{yv_{1}(x)-uw_{1}(x)}{Z(x)}, \frac{yv_{2}(x)-uw_{2}(x)}{Z(x)}\right )=0$$
Where $$W(x) = c_{1}w_{1}(x)+c_{2}w_{2}(x) , V(x) = c_{1}v_{1}(x)+c_{2}v_{2}(x)$$ is the general solution 
of the system of equations 
 $$\frac{dW}{dx}=aW+bV , \frac{dV}{dx}=cW+dV and Z=w_{1}v_{2}-w_{2}v_{1}$$
and $$Z=w_{1}v_{2}-w_{2}v_{1}.$$ Hence , solve the following partial PDEs:
$$ (i) \  u_{x}+(-y+2u)u_{y}=4y+u$$
$$ (ii)\  u_{x}+\frac{2y+u}{x}u_{y}=\frac{4y+2u}{x}$$
And the answer using part one is supposed to be as follows:
$$ (i) \  \phi (e^{-3x(y+u)},e^{3x}(u-2y))=0 $$
$$ (ii) \  \phi ((2y+u)/x^{4} ,2y-u)=0 $$
Thank You...
 A: For First Part we have :
$$\phi \left (\mu= \frac{yv_{1}(x)-uw_{1}(x)}{Z(x)},\nu= \frac{yv_{2}(x)-uw_{2}(x)}{Z(x)}\right )=0$$
by differentiating $ \phi $ w.r.to. x and y we have :
$$\phi_\mu(\mu_{x}+\mu_{u}u_{x})+\phi_\nu(\nu_{x}+\nu_{u}u_{x})=0$$
$$\phi_\mu(\mu_{y}+\mu_{u}u_{y})+\phi_\nu(\nu_{y}+\nu_{u}u_{y})=0$$
by eliminating $ \phi_\mu $ and $ \phi_\nu $ we have :
$$(\mu_{y}\nu_{u}-\mu_{u}\nu_{y})u_{x} +(\mu_{u}\nu_{x}-\mu_{x}\nu_{u})u_{y}=(\mu_{x}\nu_{y}-\mu_{y}\nu_{x})\ \ *$$
$$!\ (\mu_{y}\nu_{u}-\mu_{u}\nu_{y})=1$$
$$!!\ (\mu_{u}\nu_{x}-\mu_{x}\nu_{u})=y\left(\frac{w_{2}(x).v'_{1}(x)-w_{1}(x).v'_{2}(x)}{Z(x).Z'(x)}\right)+u\left(\frac{w_{1}(x).w'_{2}(x)-w'_{1}(x).w_{2}(x)}{Z(x).Z'(x)}\right)$$
$$!!!\ (\mu_{x}\nu_{y}-\mu_{y}\nu_{x})=y\left(\frac{v_{2}(x).v'_{1}(x)-v_{1}(x).v'_{2}(x)}{Z(x).Z'(x)}\right)+u\left(\frac{w_{2}(x).v'_{1}(x)-w'_{1}(x).v_{2}(x)}{Z(x).Z'(x)}\right)$$
substitute from ! , !! and !!! into * we have :
$$u_{x} + (a(x)y + b(x)u)u_{y}=c(x)y + d(x)u  $$
where 
$$ a(x)=\left(\frac{w_{2}(x).v'_{1}(x)-w_{1}(x).v'_{2}(x)}{Z(x).Z'(x)}\right)$$
$$b(x)=\left(\frac{w_{1}(x).w'_{2}(x)-w'_{1}(x).w_{2}(x)}{Z(x).Z'(x)}\right)$$
$$c(x)=\left(\frac{v_{2}(x).v'_{1}(x)-v_{1}(x).v'_{2}(x)}{Z(x).Z'(x)}\right)$$
$$d(x)=\left(\frac{w_{2}(x).v'_{1}(x)-w'_{1}(x).v_{2}(x)}{Z(x).Z'(x)}\right)$$
hence, the general solution of the partial DE
$$u_{x} + (a(x)y + b(x)u)u_{y}=c(x)y + d(x)u  $$
is of the form 
$$\phi \left ( \frac{yv_{1}(x)-uw_{1}(x)}{Z(x)}, \frac{yv_{2}(x)-uw_{2}(x)}{Z(x)}\right )=0$$
For Second part (i):
$$\  u_{x}+(-y+2u)u_{y}=4y+u$$
$$ a(x)=-1 , b(x)=2 ,c(x)=4, d(x)= 1$$
Now we have to solve the following system of ODEs :
$$\frac{dW}{dx}=-W+2V$$
$$ \frac{dV}{dx}=4W+1V$$
It's very easy to solve this system :
$$W=c_{1}e^{3x}-c_{2}e^{-3x}=c_{1}w_{1}(x)+c_{2}w_{2}(x)$$
$$V=2C_{1}e^{3x}+c_{2}e^{-3x}=c_{1}v_{1}(x)+c_{2}v_{2}(x)$$
So 
$$w_{1}(x)=e^{3x}$$
$$w_{2}(x)=-e^{-3x}$$
$$v_{1}(x)=2e^{3x}$$
$$v_{2}(x)=e^{-3x}$$
by substituting in the general form we have :
$$\phi\left(e^{3x}(u-2y),e^{-3x}(y+u)\right)=0 $$
Now (ii) 
$$  u_{x}+\frac{2y+u}{x}u_{y}=\frac{4y+2u}{x}$$
$$a(x)=\frac{2}{x} , b(x)=\frac{1}{x} , c(x) =\frac{4}{x} , d(x)=\frac{2}{x}
$$
we have :
$$\frac{dW}{dx}=\frac{2}{x}W+\frac{1}{x}V$$
$$ \frac{dV}{dx}=\frac{4}{x}W+\frac{2}{x}V$$
Now to solve this put $$ \frac{dx}{dt}=x \Rightarrow ln x = t $$
and 
$$\frac{dW}{dt}.\frac{dt}{dx}=\frac{2}{x}W+\frac{1}{x}V$$
$$ \frac{dV}{dx}.\frac{dt}{dx}=\frac{4}{x}W+\frac{2}{x}V$$
which implies :
$$\frac{dW}{dt}=2W+V$$
$$ \frac{dV}{dt}=4W+2V$$
by solving this system we have 
$$W=C_{1}\frac{x^{4}}{4}+\frac{1}{4}C_{2}=c_{1}w_{1}(x)+c_{2}w_{2}(x)$$
$$V=C_{1}\frac{x^{4}}{2}-\frac{1}{2}c_{2}=c_{1}v_{1}(x)+c_{2}v_{2}(x)$$
so 
$$w_{1}(x)=\frac{x^{4}}{4}$$
$$w_{2}(x)=\frac{1}{4}$$
$$v_{1}(x)=\frac{x^{4}}{2}$$
$$v_{2}(x)=\frac{-1}{2}$$
by substituting in the general form we have :
$$\phi\left(u-2y),\frac{2y+u}{x^{4}}\right)=0$$
