Ask Uncle Colin: A Separable Difficulty

Dear Uncle Colin,

I have an equation $3y,\frac{dy}{dx}=x$. When I separate and integrate both sides, I end up with $\frac{3}{2}{y}^2=\frac{1}{2}{x}^2$, which reduces to $y=x\sqrt{\frac{1}{3}}+c$.

With the initial condition $y(3)=11$, I get $y=x\sqrt{\frac{1}{3}}+11-3\sqrt{\frac{1}{3}}$, but apparently this is incorrect. What am I doing wrong?

- Getting Ridiculous Expressions, Evaluating Nonsense

Hello, GREEN, and thank you for your message!

You’ve had a good stab at this - your only problem is that left your constant of integration out for too long!

When you separate your equation, you get $\int 3y\mathrm{d}y=\int x\mathrm{d}x$, which strictly integrates to $\frac{3}{2}{y}^2+c_y=\frac{1}{2}{x}^2+c_x$, with (generally) different constants on either side. However, the two constants can be combined into one, to give $\frac{3}{2}{y}^2=\frac{1}{2}{x}^2+c$.

You could even substitute in your initial condition at this point:

$\frac{3}{2}\left({11}^2\right)=\frac{1}{2}\left(3^2\right)+c$ gives $\frac{363}{2}=\frac{9}{2}+c$, so $c=177$.

You’ve now got $\frac{3}{2}{y}^2=\frac{1}{2}{x}^2+177$, so $y^2=\frac{1}{3}{x}^2+118$, and $y=\sqrt{\frac{1}{3}{x}^2+118}$.

A method I prefer, which gets rid of the arbitrary constants, uses definite integration of dummy variables like so: ${\int }_{11}^{y}3Y\mathrm{d}Y={\int }_3^{x}X\mathrm{d}X$. This gives ${\left[\frac{3}{2}{Y}^2\right]}_{11}^y={\left[\frac{1}{2}{X}^2\right]}_3^x$; evaluating that gives $\frac{3}{2}(y^2-121)=\frac{1}{2}(x^2-9)$, leading to the same answer with a little less working.

Hope that helps!

-- Uncle Colin