3
$\begingroup$

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...

$\endgroup$
2
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.