*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'm supposed to use the change of variable $z = \sin(x)$ to turn $\cos(x) \diffn{2}{y}{x} + \sin(x) \dydx - 2y \cos^3(x)= 2\cos^5(x)$ into $\diffn{2}{y}{z} - 2y = 2\left(1-z^2\right)$.

Yeah but no but. No idea.

Lacking, Obviously, Something Trivial

Hello, LOST,

Right, yes. Nasty one, this. The main problem (to me) is that there aren't any $y$s knocking around in the substitution. However, that's easily-enough fixed if we just differentiate it with respect to $y$:

$z = \sin(x)$

$\diff {z}{y} = \cos(x) \diff {x}{y}$

Carefully using the chain rule to keep @RealityMinus3 happy:

$\diff{y}{z} = \frac{1}{\cos{x}} \times \frac{1}{\diff{x}{y}} = \sec(x) \diff{y}{x}$, which is a big step in the right direction! Now we can differentiate again to get $\diffn{2}{y}{z}$.

$\diffn{2}{y}{z} = \frac{\d}{\d z} \left( \sec(x) \diff{y}{x} \right)$

Again using the chain rule, we can treat this as an $x$-derivative as long as we multiply by $\diff {x}{z}$ afterwards -- and we know that's $\sec(x)$. Let's differentiate with respect to $x$:

$\frac{\d}{\dx} \left( \sec(x) \diff{y}{x} \right) = \sec(x) \tan(x) \diff{y}{x} + \sec(x) \diffn{2}{y}{x}$

That makes $\diffn{2}{y}{z} = \left(\sec(x) \tan(x) \diff{y}{x} + \sec(x) \diffn{2}{y}{x}\right) \sec(x)$

Tidying up slightly: $\diffn{2}{y}{z} = \left(\sec^2(x) \tan(x) \diff{y}{x} + \sec^2(x) \diffn{2}{y}{x}\right) $

The first term in our original differential equation is $\cos(x) \diffn{2}{y}{x}$, so left's divide the whole thing through by $\cos^3(x)$ to make it fit together:

$\sec^2(x) \diffn{2}{y}{x} + \tan(x)\sec^2(x) \dydx - 2y = 2\cos^2(x)$

Oh, but look! The first two terms are $\diffn{2}{y}{z}$! And $\cos^2(x)$ on the right hand side is $(1- \sin^2(x))$, or $(1-x^2)$. We're left with:

$\diffn{2}{y}{z} - 2y = 2(1-z^2)$, as required.

Phew.

-- 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.