1. ## Integralcurves and Vectorfields

Hey

i just stared learning ODE's and since have to learn it own my own i'm havin a hard time understanding!

So could you please explain to me how to calculate the integralcurves for the following vector fields:
f(x,y)=(4,6);
f(x,y)=(x,y)
f(x,y)=(x,-y)

I have no clue how to do this, i thought i should just find a function for which these are the derivatives eg. f(x,y)=(4,6)-->F(x,y)=4x+6y but i don't think its that easy, especially when i compare it with the plot (maple).

thx

2. let $\vec{r}(t)=x(t)\vec{i}+y(t)\vec{j}$ be the position vector

Then $\frac{d}{dt}\vec{r}(t)=x'(t)\vec{i}+\y'(t)\vec{j}$

using this for 1) we get

$\frac{d}{dt}\vec{r}(t)=f(\vec{r}(t) \iff x'(t)\vec{i}+\y'(t)\vec{j}=4\vec{i}+6\vec{j}$

This gives the system of ODE's

$\frac{dx}{dt}=4$ and $\frac{dy}{dt}=6$

solving gives

$x(t)=4t+c_1$ and $y(t)=6t+c_2$

Now you can eliminate t from the equations by solving one for t and plugging it into the other

$\displaystyle t=\frac{x-c_1}{4} \implies y=6\left( \frac{x-c_1}{4}\right)+c_2=\frac{3}{2}x+\left(c_2-\frac{6c_1}{4}\right)$

This is a family of lines

Can you do this for the other 2?

3. ok i tried f(x,y)=(x,y)

$x'(t)=x \rightarrow x(t)=x_0e^t$ and $y'(t)=y\rightarrow y(t)=y_0e^t$

so i solve for t and plug it into the other equation:

$t=ln(\frac{x}{x_0})\rightarrow y(t)=y_0e^{ln(\frac{x}{x_0})}=y(t)=\frac{y_0}{x_0} x$

is that correct?