*Ask Uncle Colin* is a chance to ask your burning, possibly embarrassing, maths questions -- and to show off your skills at coming up with clever acronyms. Send your questions to colin@flyingcoloursmaths.co.uk and Uncle Colin will do what he can.

Dear Uncle Colin,

I have an equation $3y, \dydx =x$. 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 $\int 3y \d y = \int x \dx$, 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: $\int_{11}^y 3Y \d{Y} = \int_3^x X\d{X}$. 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

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008.
He lives with an espresso pot and nothing to prove.

## Stephen Cavadino

I love the definite integration method. I think it makes a lot more sense.