1. ## Vector Fields

I don't think this problem is too difficult, but I have no idea for what it's asking:

For the vector field $\textbf{F}(x,y)=\textbf{i}+x\textbf{j}$, if the parametric flow lines are $x=x(t)$, $y=y(t)$, what differential equations do these functions satisfy? Deduce that $\frac{dy}{dx}=x$.

First off, I don't know what $x=x(t)$ and $y=y(t)$ mean, and how $t$ comes into play. Could someone help?

2. Originally Posted by MathTooHard
I don't think this problem is too difficult, but I have no idea for what it's asking:

For the vector field $\textbf{F}(x,y)=\textbf{i}+x\textbf{j}$, if the parametric flow lines are $x=x(t)$, $y=y(t)$, what differential equations do these functions satisfy? Deduce that $\frac{dy}{dx}=x$.

First off, I don't know what $x=x(t)$ and $y=y(t)$ mean, and how $t$ comes into play. Could someone help?

$x = x(t) , y= y(t)$ is/are the parametric equation(s) which you are looking for . $t$ is the parameter , not necessary to be the time , we can replace it with a,b,c,d,e,f,g any letter you like .

Usually , we could find out the parametric equation AFTER we find out the function(family) of the lines $y = f(x)$ , so don't worry about $x(t)$ and $y(t)$

3. Are you asked to solve the differential equation ?

for the flow lines $y = f(x)$in a vector field $F(x,y) = P(x,y)i + Q(x,y)j$

we always have $\frac{dy}{dx} = f'(x) = \frac{Q(x,y) }{P(x,y)}$

4. Originally Posted by simplependulum
Are you asked to solve the differential equation ? Nope.

for the flow lines $y = f(x)$in a vector field $F(x,y) = P(x,y)i + Q(x,y)j$

we always have $\frac{dy}{dx} = f'(x) = \frac{Q(x,y) }{P(x,y)}$
From that, I get $\frac{dy}{dx} = \frac{x}{1}= x$. But how should I go about figuring out what differential equations x=x(t) and y=y(t) satisfy? I don't really know what a step of t should mean in such a parametrization. I think x'(t)=1, but what would y'(t) be?