Solving a Flow Line Curve: c(t) for F=(x,-y)

[madachi]

Homework Statement

Find the flow line curve $c(t)$ to the vector field $F = (x,-y)$ which passes through the point $(1, 2)$.

The Attempt at a Solution

So I let $c(t) = (x(t), y(t))$.
So $c'(t) = ( \frac{dx}{dt} , \frac{dy}{dt} )$.

Now, $\frac{dx}{dt} = x$ and $\frac{dy}{dt} = -y$.

So $\frac{dy}{dx} = -\frac{y}{x}$

Solving the differential equation, I get

$ln(y) = -ln(x) + C$
$y = e^{-ln(x) + C}$
$y = \frac{A}{x}$
$y = \frac{2}{x}$ by using the point given.

This is not the answer given, I am not sure what they want. The answer given is
$c(t) = ( e^{t}, 2e^{-t} )$.

Thanks.

 

[lanedance]

you have given an explicit equation y(x).

The answer given is parametric c(t) = (y(t), x(t))

 

[lanedance]

there's an arithmetic mistep as well...

1. Homework Statement


Find the flow line curve $c(t)$ to the vector field $F = (x,-y)$ which passes through the point $(1, 2)$.

The Attempt at a Solution

So I let $c(t) = (x(t), y(t))$.
So $c'(t) = ( \frac{dx}{dt} , \frac{dy}{dt} )$.

Now, $\frac{dx}{dt} = x$ and $\frac{dy}{dt} = -y$.

so you could solve for the parametric form of y(t) and x(t) here rather than the explicit substitution

use the given point as you initial conditions for each

So $\frac{dy}{dx} = -\frac{y}{x}$

Solving the differential equation, I get

$ln(y) = -ln(x) + C$
$y = e^{-ln(x) + C}$

the next step isn't quite right either, it should go
$y = e^{-ln(x) + C} = e^C e^{ln(x^{-1})} = e^C (x^{-1})$

$y = Ax$
$y = 2x$ by using the point given.

This is not the answer given, I am not sure what they want. The answer given is
$c(t) = ( e^{t}, 2e^{-t} )$.

Thanks.