# 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...

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 : $$\phi\left(u-2y),\frac{2y+u}{x^{4}}\right)=0$$