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Let $\textbf{F}:\mathbb{R}^2 \to \mathbb{R}^2$ be a vector field defined as \begin{equation} \textbf{F}(x,y) = (x^2, y+1) \end{equation} Find the streamline of $\textbf{F}$ that passes through the point $(1,1)$.

Answer

Define \begin{equation} \textbf{r}(t) = (x(t), y(t)) \end{equation} then \begin{equation} \textbf{r}'(t) = (x'(t), y'(t)) = \textbf{F}(x(t),y(t)) = (x^2(t), y(t)+1) \end{equation} so we get two differential equations. Namely \begin{equation} x'(t) = x^2(t) \mbox{ and } y'(t) = y(t) + 1 \end{equation} We beging solving \begin{align} y(t) = e^{t} \int e^{-t} dy = e^{t} (-e^{-t} + C) = e^{t}C - 1 \end{align} and \begin{equation} \int \frac{\frac{\partial x(t)}{\partial t}}{x^2(t)} = \int 1 \partial t \end{equation} so \begin{equation} x(t)= -\frac{1}{t + B} \end{equation} Now \begin{equation} \textbf{r}(0) = (x(0), y(0)) = (-\frac{1}{t + B}, e^{t}C - 1) = (1,1) \end{equation} So $C = 2$ and $B = -1$ and \begin{equation} \textbf{r}(t) = (x(t), y(t)) = (-\frac{1}{t - 1}, 2e^{t} - 1) \end{equation}

Is this correct?

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In the last step, when $t=0$, you should have $({-1\over 0+B}, e^0C-1)=(1,1)$. So $C=2, B=-1$ and $r(t)=({-1\over t-1}, 2e^t-1)$. –  David Mitra Mar 8 '12 at 13:37
    
Whoops. Thanks! –  docjay Mar 8 '12 at 13:38
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up vote 1 down vote accepted

It looks fine except for the last step where you're solving for the constants. When $t=0$, you should have $${\bf r}(0)=\bigl(\,\textstyle{−1\over 0+B},e^0\cdot C−1\,\bigr)=(1,1)\ \ \Rightarrow \ \ B=−1,\ C=2 \ \ \Rightarrow\ \ {\bf r}(t)=\bigl(\,{-1\over t-1},2e^t−1\,\bigr). $$

And on a picky note, at the outset, you should define ${\bf r}(t)=\bigl(x(t),y(t)\bigr)$, ${\bf r}(0)=(1,1)$.

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