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Math Help - Integralcurves and Vectorfields

  1. #1
    Junior Member
    Joined
    Oct 2010
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    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).

    Can you help me please?

    thx
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  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Yuma, AZ, USA
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    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?
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  3. #3
    Junior Member
    Joined
    Oct 2010
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    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?
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