Reduce the Cauchy-Euler equation to a smaller expression?

Question

Given the Cauchy-Euler second order differential equation: $$ \left( x^2\frac{d^2}{d x^2} + x\frac{d}{d x} -1\right)y(x)=0 $$ is it possible to find a function $u(x)$ , which satisfies the following simple form:? $$ \frac{d^2 y(u)}{d u^2}=0 $$ ***notice that $t=\ln(x)$ yields: $$ \frac{d^2 y(t)}{d t^2}-1=0 $$

Answer 1

notice that t=ln(x) yields: $$\dfrac {d^2y(t)}{dt^2}−1=0$$

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$$ \left( x^2\frac{d^2}{d x^2} + x\frac{d}{d x} -1\right)y(x)=0$$ It yields: $$\dfrac {d^2y(t)}{dt^2}−\color {red}{y(t)}=0$$ That you can rewrite as: $$y''+y'-y'-y=0$$ $$(ze^{-t})'=0$$ Where $z=y'+y$. $$((ye^t)'e^{-2t})'=0$$

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Note that: $$\frac{d^2 y(u)}{d u^2}=0$$ Has solution: $$y(u)=c_1u+c_2$$ But $c_2$ is not a solution of the original equation: $$ x^2\frac{d^2y}{d x^2} + x\frac{dy}{d x} -y=0$$ So it 's not possible to find such $u=u(x)$.

Answer 2

HINT

To begin with, notice that

\begin{align*} x^{2}y'' + xy' - y = 0 & \Longleftrightarrow (x^{2}y'' + 2xy') - (xy' + y) = 0\\\\ & \Longleftrightarrow (x^{2}y')' - (xy)' = 0\\\\ & \Longleftrightarrow x^{2}y' - xy = c_{0} \end{align*}

Can you take it from here?

Answer 3

we suppose that there exits a $u(x)$, if so; using chain rule we have: $$ \frac{d^2y}{du^2} = \left(\frac{dx}{du}\right)^2 \frac{d^2y}{dx^2} + \frac{dx}{du} \frac{d\left(\frac{dx}{du}\right)}{dx} \frac {dy}{dx}$$ now setting this equal to the main equation: $$ x^2\frac{d^2y}{dx^2} + x\frac{dy}{dx} -y $$ there is an extra term in the main equation ($-y$) which cannot be found. however if this extra term didnot exist, we'd have: $$ \left(\frac{dx}{du}\right)^2 \frac{d^2y}{dx^2} = x^2\frac{d^2y}{dx^2} $$ $$ \frac{dx}{du} \frac{d\left(\frac{dx}{du}\right)}{dx} \frac {dy}{dx} = x\frac{dy}{dx} $$ these two equations are dependant, i.e. a solution to one of them is a solution to the other. therefore solving the first one suggests: $$ u = \ln(x) $$ this is valid for the second one, too.

Answer 4

Starting backwards from the solution $y=c_1x+c_2x^{-1}$, if you set $xy(x)=f(x^2)$, then you get for the derivatives $$ xy'+y=2xf'(x^2)\\ xy''+2y'=4x^2f''(x^2)+2f'(x^2) $$ so that $$ 0=x^2y''+xy'-y=4x^3f''(x^2). $$ This gives the expected equation $f''=0$, but not exactly the way you were looking for.