I need help with the following question: The given equation is:


Show that the characteristic lines of the equation are: $$\begin{cases} x(t)=C_1e^t+C_2e^{-t}&\\ y(t)=C_1e^t-C_2e^{-t}&\\ u(t)=\frac{1}{C_3-t} &&or&& u(t)\equiv0 \end{cases}$$ My idea is to solve the Following ODE's system: $$\begin{cases} x_t=y&\\ y_t=x&\\ u_t=u^2\end{cases}$$ i still not figure out how i need to to that.that's not a classic ODE. tried to integrae

$x_t=y$ and $y_t=x$ separately but then i don't get the right answer. tried to solve ODE first-order system of X and Y but I found that the eigenvalue is equal to 0.

Any ideas? thanks:)


Write ( I use s instead of t) : $$\frac {dx}{ds}=y$$ Differentiate wrt s: $$\frac {d^2x}{ds^2}=\frac {dy}{ds}$$ Note that we have $$\frac {dy}{ds}=x$$ So that we have, $$\frac {d^2x}{ds^2}-x=0$$ It's linear of second order

$$r^2-1=0 \implies r=\pm 1 \implies x(s)=c_1e^s+c_2e^{-s}$$ for $y(s)$ $$\frac {dx}{ds}=y \implies y(s)=c_1e^s-c_2e^{-s}$$

For the last one

For $u=0$ as @holo pointed out in the comment we have $u=c$ And for $u \ne 0$ we have that $$\frac {du}{ds}=u^2 \implies \int \frac {du}{u^2}=\int ds$$ $$\implies u(s)=\frac 1 {c_3-s}$$

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    $\begingroup$ In the last line you should note that there exists constant solution($u\equiv 0$) $\endgroup$ – ℋolo Oct 4 '18 at 16:02
  • $\begingroup$ Oh yes @holo thanks a lot $\endgroup$ – Isham Oct 4 '18 at 16:05
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    $\begingroup$ @Holo Corrected thank you $\endgroup$ – Isham Oct 4 '18 at 16:07
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    $\begingroup$ while finding $y(s)$ we know that $y_t=x$ and we found x so we can just integrate both sides with respect to s. and get the answer.we get that$C_1=C_4,-C_2=C_5$ Thanks:) $\endgroup$ – Omer Ben Oct 4 '18 at 17:50
  • $\begingroup$ @OmerBen you're welcome...yes thats correct for the coefficients $\endgroup$ – Isham Oct 4 '18 at 18:19

You can see that $$ x_t+y_t=x+y $$ and $$ y_t-x_t=-(y-x) $$ which are both scalar equations that lead to the solution of the first two equation.

Or you could consider that $$ x_{tt}=y_t=x $$ which again gives the given solution for $x$ and $y=x_t$.

Choosing the way of the Lagrange equations $$ \frac{dx}y=\frac{dy}x=\frac{du}{u^2} $$ results in the constants along characteristics $y^2-x^2=c_1$ and $\ln|x+y|+\frac1u=c_2$ which are dependent, $c_2=\Phi(c_1)$, which implies the general solution form $$ u=\frac1{\Phi(y^2-x^2)-\ln|x+y|}. $$


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