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I'm having a problem to figure out what I'm doing is correct or not. I tried to find the streamline equation with a method I was taught by the textbook and vs another method I found on youtube by Simmy Sigma.

$$\vec{v}=v_x\vec{i}+v_y\vec{j}=xy\vec{i}+y\vec{j}$$

What I did:

$$u=xy,v=y$$

$$udy=ydx$$

which gives us

$$\int{dy}=\int{\frac{1}{x}}dx$$

$$\boxed{y=ln(x)+C}$$.

then I was looking for guides on youtube and stumbled across another method which does the following.

$$d\lambda=\frac{dx}{v_x}=\frac{dy}{v_y}$$

$$(\frac{dx}{xy}=\frac{dy}{y})xy\Longrightarrow dx=xdy$$

$$-\int{xdx}=\int{dy}$$

$$-\frac{x^2}{2}=y+C$$

$$\boxed{x=\sqrt{(-2y)+(-2C)}}$$

I'm confused to which of the answer is right, if one of them is even correct.If I was to draw this streamline by hand. I just change the value of $C=0,1,2...$ etc?

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  • $\begingroup$ Shouldn't $dx = xdy$ go to $\int dy = \int \frac{1}{x} dx$? $\endgroup$ – Beta Decay Feb 25 '18 at 14:37
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Your error occurs between here:$$dx=xdy$$ and here: $$-\int{xdx}=\int{dy}$$

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  • $\begingroup$ is the first approach correct? $\endgroup$ – Sami Shafi Feb 25 '18 at 15:58
  • $\begingroup$ Yes. The two results should match. $\endgroup$ – Chet Miller Feb 26 '18 at 0:05

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