Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $(1,1)$.


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?

share|cite|improve this question
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
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 ${\bf r}(t)=\bigl(x(t),y(t)\bigr)$, ${\bf r}(0)=(1,1)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.