Ask Uncle Colin: Changing the variable (FP2 Differential equations)

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 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.


-- Uncle 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.


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