Prove that $(1−x)^2\frac{d^2y}{dx^2}−x\frac{dy}{dx}+p^2y=0$

Question

Prove that $$(1−x)^2\frac{d^2y}{dx^2}−x\frac{dy}{dx}+p^2y=0 $$

where $$y = \sin(pt)$$ and $$ x =\sin(t) $$

please, tell me how this makes sense, shouldn't the second derivative be $0$?

I am wrestling with this every day for about a month and can't make sense of it.

Answer 1

The difficulty is to make sense of $dy/dx$ when $x$ and $y$ are both parameterized by $t$. One approach would be to write $t = \arcsin x$, so $y = \sin(p\arcsin x)$. This makes for an unwieldy expression to differentiate twice.

Another approach is to write $$dy/dx = \frac{dy/dt}{dx/dt} = \frac{p\cos pt}{\cos t}.$$ Then you would have $$d^2y/dx^2 = \frac{d\left( \frac{dy/dt}{dx/dt} \right)/dt}{dx/dt}$$ $$=\frac{d\left( \frac{p\cos pt}{\cos t} \right)/dt}{\cos t}.$$ You should be able to continue from here using the quotient rule. Then you can use the computed expressions to verify the original equation.

Answer 2

$$\frac{dy}{dx}=\frac{dy}{dt}/\frac{dx}{dt}=\frac{p\cos(pt)}{\cos(t)}$$

and this does not have zero derivative in the way that was worrying you.

Answer 3

There is some abuse of notation involved, as $y$ is defined as a function of $t$, but its derivatives are written as if it were a function of $x$. To resolve this, write $y$ as composite function $y(x(t))$. There is no need to get more specific than this.

You can use that by the chain rule $$ \frac{dy(x(t))}{dt}=\frac{dy(x(t))}{dx}\frac{dx(t)}{dt} $$ and in the second derivative applying chain and product rule $$ \frac{d^2y(x(t))}{dt^2}=\frac{d^2y(x(t))}{dx^2}\left(\frac{dx(t)}{dt}\right)^2+\frac{dy(x(t))}{dx}\frac{d^2x(t)}{dt^2}. $$ Now we can insert $y(x(t))=\sin(pt)$ etc. to get the derivatives in the equation, \begin{align} p\cos(pt)&=\frac{dy}{dx}\cos(t)\\ -p^2\sin(pt)&=\frac{d^2y}{dx^2}\cos^2(t)-\frac{dx}{dx}\sin(t) \end{align} However, inserting the trigonometric identity $\sin^2t+\cos^2t=1$ into the last equation and inserting back $\sin(t)=x$ etc. we already get the differential equation for $y(x)$, $$ (1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}+p^2y=0. $$

Answer 4

The relation you are trying to prove is incorrect.

Assuming the relation you meant to prove was $$ (1-x^2) \frac{d^2y}{dx^2}-x\frac{dy}{dx}+p^2y = 0 $$ here is what you do:

$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{p\cos(pt)}{\cos t} \\ x \frac{dy}{dx} = p \cos(pt) \tan t $$

$$ \frac{d^2y}{dx^2} = \frac{d}{dx} \left(\frac{p\cos(pt)}{\cos t} \right) = \frac{1}{dx/dt} \frac{d}{dt} \frac{p\cos(pt)}{\cos t} = \frac{1}{\cos t} \left( \frac{ -p^2\sin(pt)\cos t+p\cos(pt)\sin t}{\cos^2 t}\right) \\ (1-x^2)\frac{d^2y}{dx^2} =(1-\sin^2 t) \frac{d^2y}{dx^2} = \cos^2t \frac{d^2y}{dx^2} = \frac{ -p^2\sin(pt)\cos t+p\cos(pt)\sin t}{\cos t} \\ (1-x^2)\frac{d^2y}{dx^2} = -p^2\sin(pt) + p\cos(pt) \tan t $$ Add these together to get $$ (1-x^2) \frac{d^2y}{dx^2}-x\frac{dy}{dx}+p^2y = -p^2\sin(pt) + p\cos(pt) \tan t +p \cos(pt) \tan t + p^2 y = 0 $$ since $y = \sin t$.