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.