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question : find the general solution of $(y+ux)u_x+(x+yu)u_y=u^2-1$

$\frac{dx}{dt}=y+ux,\frac{dy}{dt}=x+yu,\frac{du}{dt}=u^2-1$ I dont know how to start. is this quasilinear ?

edit 1: tried

$\frac{\frac{dx}{dt}}{y+ux}=\frac{\frac{dy}{dt}}{x+yu}=\frac{\frac{du}{dt}}{u^2-1}$

$\frac{\frac{dx}{dt}+\frac{dy}{dt}}{x+y+ux+uy}=\frac{\frac{du}{dt}}{u^2-1} $

$\frac{\frac{d(y+x)}{dt}}{x+y}=\frac{\frac{du}{dt}}{u-1}$

$\frac{d(ln(x+y)+ln(u-1))}{dt}=0$ so we have a sltion whose derivative is 0. we need to find another independent solution to optain j(x,y,u)=F(h(x,y,u))

$\frac{\frac{dx}{dt}-\frac{dy}{dt}}{y-x+ux-uy}=\frac{\frac{du}{dt}}{u^2-1} $

$\frac{\frac{d(x-y)}{dt}}{y-x}=\frac{\frac{du}{dt}}{u+1}$

$\frac{d(-ln(x-y)+ln(u+1))}{dt}=0$

hence $ln(x+y) +ln(u-1)=c_1$and $ln(u+1)-ln(x-y)=c_2$ so the solution is in implct from

$ln(x+y) +ln(u-1)=F(ln(u+1)-ln(x-y))$
F is arbitrary is this approach correct?

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  • $\begingroup$ It is quasilinear. $\endgroup$ – Chris K Apr 3 '14 at 1:54
  • $\begingroup$ A related technique. $\endgroup$ – Mhenni Benghorbal Apr 3 '14 at 2:11
  • $\begingroup$ @ChrisK edited the question can u check again please $\endgroup$ – lyme Apr 3 '14 at 3:11
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Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{du}{dt}=u^2-1$ , letting $u(0)=0$ , we have $u=-\tanh t$

$\therefore\begin{cases}\dfrac{dx}{dt}=y-x\tanh t~......(1)\\\dfrac{dy}{dt}=x-y\tanh t~......(2)\end{cases}$

$(1)+(2)$ :

$\dfrac{dx}{dt}+\dfrac{dy}{dt}=x+y-(x+y)\tanh t$

$\dfrac{d(x+y)}{dt}=(x+y)(1-\tanh t)$

$\dfrac{d(x+y)}{x+y}=(1-\tanh t)~dt$

$\int\dfrac{d(x+y)}{x+y}=\int(1-\tanh t)~dt$

$\ln(x+y)=t-\ln\cosh t+c_1$

$x+y=C_1e^t~\text{sech}~t$

$(1)-(2)$ :

$\dfrac{dx}{dt}-\dfrac{dy}{dt}=y-x-(x-y)\tanh t$

$\dfrac{d(x-y)}{dt}=(x-y)(-1-\tanh t)$

$\dfrac{d(x-y)}{x-y}=(-1-\tanh t)~dt$

$\int\dfrac{d(x-y)}{x-y}=\int(-1-\tanh t)~dt$

$\ln(x-y)=-t-\ln\cosh t+c_2$

$x-y=C_2e^{-t}~\text{sech}~t$

$\therefore\begin{cases}x=\dfrac{C_1e^t~\text{sech}~t+C_2e^{-t}~\text{sech}~t}{2}\\y=\dfrac{C_1e^t~\text{sech}~t-C_2e^{-t}~\text{sech}~t}{2}\end{cases}$

$x(0)=x_0$ , $y(0)=f(x_0)$ :

$\begin{cases}\dfrac{C_1+C_2}{2}=x_0\\\dfrac{C_1-C_2}{2}=f(x_0)\end{cases}$

$\begin{cases}C_1=x_0+f(x_0)\\C_2=x_0-f(x_0)\end{cases}$

$\therefore\begin{cases}x=\dfrac{(x_0+f(x_0))e^t~\text{sech}~t+(x_0-f(x_0))e^{-t}~\text{sech}~t}{2}\\y=\dfrac{(x_0+f(x_0))e^t~\text{sech}~t-(x_0-f(x_0))e^{-t}~\text{sech}~t}{2}\end{cases}$

$\begin{cases}x=x_0+f(x_0)\tanh t\\y=x_0\tanh t+f(x_0)\end{cases}$

$\therefore\begin{cases}x_0=\dfrac{x-y\tanh t}{1-\tanh^2t}=\dfrac{x+yu}{1-u^2}\\f(x_0)=\dfrac{y-x\tanh t}{1-\tanh^2t}=\dfrac{xu+y}{1-u^2}\end{cases}$

Hence $\dfrac{xu+y}{1-u^2}=f\left(\dfrac{x+yu}{1-u^2}\right)$

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