Finding integral curves of a vector field

[CptXray]

Homework Statement

For a vector field $$\begin{equation}
X:=y\frac{\partial{}}{\partial{x}} + x\frac{\partial{}}{\partial{y}}
\end{equation}$$
Find it's integral curves and the curve that intersects point $$p = \left(1, 0 \right).$$
Show that $$X(x,y)$$ is tangent to the family of curves: $$x^2 - y^2 = k,k∈ℝ$$

Homework Equations

The Attempt at a Solution

I know that a integral curve here is:
$$

\begin{bmatrix}
\dot{x} \\
\dot{y}\\
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 \\
1 & 0\\
\end{bmatrix}

\begin{bmatrix}
x(t)\\
y(t)\\
\end{bmatrix}$$
Solving these gives me:
$$
\begin{cases}
x(t) = yt + x_{0}
& \\
y(t) = xt + y{0}
\end{cases}
$$
For point (1, 0):
$$
\begin{cases}
x(0) = 0 \rightarrow x_{0} = 1
& \\
y(0) = 0 \rightarrow y_{0} = 0
\end{cases}
$$
I guess that's what I was supposed to do here but i can't find a way to prove that $$x^2 - y^2 = k
$$
I'd be glad for help because I couldn't find anything helpful in my textbooks.

P.S.
Hello people, I'm new and happy to find this place :)

 

[andrewkirk]

Solving these gives me:
$$
\begin{cases}
x(t) = yt + x_{0}
& \\
y(t) = xt + y{0}
\end{cases}
$$

I don't think that can be correct because it leads to a parametrisation $x=\frac1{1-t^2},y= \frac t{1-t^2}$, and that doesn't satisfy the hyperbola equation $x^2-y^2=1$.

I'm not very good at integration but fortunately we don't need to integrate. They've told us that the result satisfies $x^2-y^2=k$ and that it goes through (1,0), from which we can infer that $k=1$. So now we just need to find a parametrisation of that hyperbola and check that it satisfies the original DEs.

There are enough search terms in the above that an answer could be found by internet search. But more challenging and instructive is to note that the equation $x^2-y^2=1$ looks like a trig identity where $x$ and $y$ are trig functions of some parameter $t$. Can you think of a trig identity that has that general form?

If you can follow that path to get a parametrisation, you then just need to check it satisfies the given DEs.

 

[Ray Vickson]

Homework Statement

For a vector field $$\begin{equation}
X:=y\frac{\partial{}}{\partial{x}} + x\frac{\partial{}}{\partial{y}}
\end{equation}$$
Find it's integral curves and the curve that intersects point $$p = \left(1, 0 \right).$$
Show that $$X(x,y)$$ is tangent to the family of curves: $$x^2 - y^2 = k,k∈ℝ$$

Homework Equations

The Attempt at a Solution

I know that a integral curve here is:
$$

\begin{bmatrix}
\dot{x} \\
\dot{y}\\
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 \\
1 & 0\\
\end{bmatrix}

\begin{bmatrix}
x(t)\\
y(t)\\
\end{bmatrix}$$
Solving these gives me:
$$
\begin{cases}
x(t) = yt + x_{0}
& \\
y(t) = xt + y{0}
\end{cases}
$$
Hello people, I'm new and happy to find this place :)

No, you cannot solve the coupled DEs the way you did. The DEs read as
$$\begin{array}{rcl}
\dot{x}(t) &=& y(t) \\
\dot{y}(t) & =& x(t)
\end{array}$$ You cannot just erase the "$(t)$" part of $y(t)$ and then declare that $x(t) = x_0 + y t.$ The actual solution $(x(t),y(t))$ is not given by a pair of linear functions of $t$.

 

[CptXray]

I think I've found general solution:
$$
\dot{\gamma}(t) = \alpha \begin{pmatrix} 1 \\ 1 \end{pmatrix} e^{t} + \beta \begin{pmatrix} 1 \\ -1 \end{pmatrix} e^{-t},
$$
where
$$
\alpha,\beta = const.
$$
If anyone doesn't mind I'll upload detailed solution step by step later because it's really late here in my time zone.